Let's say $A$ is a $6\times6$ matrix with characteristic polynomial $p(x)=(x-\lambda_1)^2(x-\lambda_2)^4$, for example, when $\lambda_1\neq\lambda_2$. I have a doubt about obtaining all possibilities for the Jordan form of $A$:
If the eigenspaces correspondent to $\lambda_1$ and $\lambda_2$ are 1 and 2-dimensional, respectively, then the only possibility excluded from all the JCF is the diagonal form
$$\begin{pmatrix}\lambda_1 & 0 & 0 & 0 & 0 & 0\\ 0 & \lambda_1 & 0 & 0 & 0 & 0\\ 0 & 0 & \lambda_2 & 0 & 0 & 0\\ 0 & 0 & 0 &\lambda_2 & 0 & 0\\ 0 & 0 & 0 & 0 & \lambda_2 & 0\\ 0 & 0 & 0 & 0 & 0 & \lambda_2 \end{pmatrix}$$ because otherwise the correspondent eigenvalues wouldn't have dimension 1 and 2, correct?
Further, let's say I know the minimal polynomial to be $m(x)=(x-\lambda_1)(x-\lambda_2)^4$, for example. Which other possibilities from the JFC can I remove in this case?
If the eigenspaces for $\lambda_1$ and $\lambda_2$ are $1$- and $2$-dimensional, respectively, then you exclude all Jordan forms in which there is more than one block corresponding to $\lambda_1$, or not exactly two blocks corresponding to $\lambda_2$. This because the dimension of the eigenspace equals the number of Jordan blocks associated to that eigenvalue. So either there are two blocks associated to $\lambda_1$, or there is exactly $1$, $3$, or $4$ blocks associated to $\lambda_2$. In addition to the diagonal form (where you have two $\lambda_1$ blocks and four $\lambda_2$ blocks), you also exclude the following forms: $$\begin{align*} \left(\begin{array}{cccccc} \lambda_1 & 0 & 0 & 0 & 0 & 0 \\ 0 & \lambda_1 & 0 & 0 & 0 & 0 \\ 0 & 0 & \lambda_2 & 1 & 0 & 0\\ 0 & 0 & 0 & \lambda_2 & 0 & 0\\ 0 & 0 & 0 & 0 & \lambda_2 & 0\\ 0 & 0 & 0 & 0 & 0 & \lambda_2 \end{array}\right),&\qquad\text{two $\lambda_1$ blocks and three $\lambda_2$ blocks};\\\ \strut\\ \left(\begin{array}{cccccc} \lambda_1 & 0 & 0 & 0 & 0 & 0 \\ 0 & \lambda_1 & 0 & 0 & 0 & 0 \\ 0 & 0 & \lambda_2 & 1 & 0 & 0\\ 0 & 0 & 0 & \lambda_2 & 1 & 0\\ 0 & 0 & 0 & 0 & \lambda_2 & 0\\ 0 & 0 & 0 & 0 & 0 & \lambda_2 \end{array}\right),&\qquad\text{two $\lambda_1$ blocks and two $\lambda_2$ blocks};\\ \strut\\ \left(\begin{array}{cccccc} \lambda_1 & 0 & 0 & 0 & 0 & 0 \\ 0 & \lambda_1 & 0 & 0 & 0 & 0 \\ 0 & 0 & \lambda_2 & 1 & 0 & 0\\ 0 & 0 & 0 & \lambda_2 & 0 & 0\\ 0 & 0 & 0 & 0 & \lambda_2 & 1\\ 0 & 0 & 0 & 0 & 0 & \lambda_2 \end{array}\right),&\qquad\text{two $\lambda_1$ blocks and two $\lambda_2$ blocks};\\ \strut\\ \left(\begin{array}{cccccc} \lambda_1 & 0 & 0 & 0 & 0 & 0 \\ 0 & \lambda_1 & 0 & 0 & 0 & 0 \\ 0 & 0 & \lambda_2 & 1 & 0 & 0\\ 0 & 0 & 0 & \lambda_2 & 1 & 0\\ 0 & 0 & 0 & 0 & \lambda_2 & 1\\ 0 & 0 & 0 & 0 & 0 & \lambda_2 \end{array}\right),&\qquad\text{two $\lambda_1$ blocks and one $\lambda_2$ block};\\ \strut\\ \left(\begin{array}{cccccc} \lambda_1 & 1 & 0 & 0 & 0 & 0 \\ 0 & \lambda_1 & 0 & 0 & 0 & 0 \\ 0 & 0 & \lambda_2 & 0 & 0 & 0\\ 0 & 0 & 0 & \lambda_2 & 0 & 0\\ 0 & 0 & 0 & 0 & \lambda_2 & 0\\ 0 & 0 & 0 & 0 & 0 & \lambda_2 \end{array}\right), &\qquad\text{ four $\lambda_2$ blocks};\\ \strut\\ \left(\begin{array}{cccccc} \lambda_1 & 1 & 0 & 0 & 0 & 0 \\ 0 & \lambda_1 & 0 & 0 & 0 & 0 \\ 0 & 0 & \lambda_2 & 1 & 0 & 0\\ 0 & 0 & 0 & \lambda_2 & 0 & 0\\ 0 & 0 & 0 & 0 & \lambda_2 & 0\\ 0 & 0 & 0 & 0 & 0 & \lambda_2 \end{array}\right),&\qquad\text{three $\lambda_2$ blocks};\\ \strut\\ \left(\begin{array}{cccccc} \lambda_1 & 1 & 0 & 0 & 0 & 0 \\ 0 & \lambda_1 & 0 & 0 & 0 & 0 \\ 0 & 0 & \lambda_2 & 1 & 0 & 0\\ 0 & 0 & 0 & \lambda_2 & 1 & 0\\ 0 & 0 & 0 & 0 & \lambda_2 & 1\\ 0 & 0 & 0 & 0 & 0 & \lambda_2 \end{array}\right),&\qquad\text{one $\lambda_2$ block}. \end{align*}$$
If you know that the characteristic polynomial is $(x-\lambda_1)^2(x-\lambda_2)^4$ and the dimension of the eigenspaces corresponding to $\lambda_1$ and $\lambda_2$ are, respectively, $1$ and $2$, then there are two possible Jordan Canonical Forms: $$\left(\begin{array}{cccccc} \lambda_1 & 1 & 0 & 0 & 0 & 0 \\ 0 & \lambda_1 & 0 & 0 & 0 & 0 \\ 0 & 0 & \lambda_2 & 1 & 0 & 0\\ 0 & 0 & 0 & \lambda_2 & 1 & 0\\ 0 & 0 & 0 & 0 & \lambda_2 & 0\\ 0 & 0 & 0 & 0 & 0 & \lambda_2 \end{array}\right)\qquad\text{or}\qquad \left(\begin{array}{cccccc} \lambda_1 & 1 & 0 & 0 & 0 & 0 \\ 0 & \lambda_1 & 0 & 0 & 0 & 0 \\ 0 & 0 & \lambda_2 & 1 & 0 & 0\\ 0 & 0 & 0 & \lambda_2 & 0 & 0\\ 0 & 0 & 0 & 0 & \lambda_2 & 1\\ 0 & 0 & 0 & 0 & 0 & \lambda_2 \end{array}\right);$$ as these are the only forms that have exactly one $\lambda_1$-block and exactly two $\lambda_2$-blocks. The minimal polynomials would be $(x-\lambda_1)^2(x-\lambda_2)^3$ and $(x-\lambda_1)^2(x-\lambda_2)^2$, respectively.
The degree of the factor $(x-\lambda)$ in the minimal polynomial is the size of the largest Jordan block associated to $\lambda$. If the minimal polynomial is $(x-\lambda_1)(x-\lambda_2)^4$, then that tells you that the largest block corresponding to $\lambda_1$ is $1\times 1$, and the largest block corresponding to $\lambda_2$ is $4\times 4$. A matrix with characteristic polynomial $(x-\lambda_1)^2(x-\lambda_2)^4$ and minimal polynomial $(x-\lambda_1)(x-\lambda_2)^4$ is necessarily similar to the matrix $$\left(\begin{array}{cccccc} \lambda_1 & 0 & 0 & 0 & 0 & 0 \\ 0 & \lambda_1 & 0 & 0 & 0 & 0 \\ 0 & 0 & \lambda_2 & 1 & 0 & 0\\ 0 & 0 & 0 & \lambda_2 & 1 & 0\\ 0 & 0 & 0 & 0 & \lambda_2 & 1\\ 0 & 0 & 0 & 0 & 0 & \lambda_2 \end{array}\right).$$ Note, however, that this is incompatible with the dimension of the eigenspace corresponding to $\lambda_2$ being $2$ and with the dimension of the eigenspace corresponding to $\lambda_1$ being $1$. So the two conditions you state ((i) the dimensions of the eigenspaces, and (ii) the minimal polynomial) are incompatible with one another.
In general, the minimal and characteristic polynomials may not suffice to completely determine the Jordan Canonical Form (once you get past dimension $3$, there are always examples where they don't uniquely determine the JCF). Adding the dimension of the eigenspaces will also not suffice in general. For instance, there are two different JCFs for a $7\times 7$ matrix with characteristic polynomial $(x-\lambda)^7$, minimal polynomial $(x-\lambda)^3$, and with the eigenspace of $\lambda$ having dimension $3$: one with two $3\times 3$ blocks and one $1\times 1$ block; and one with one $3\times 3$ block and two $2\times 2$ blocks.