If $A$ is a matrix s.t. $det(A)\neq0$ mod $p$ and $\vec x$ is a primitive vector, i.e. not all its entries are divisible by $p$ then $A\vec{x}\neq\vec 0$ mod $p$.
In the normal case, without reducing it modulo $p$ it is clear that $A\vec x\neq\vec0$ since $A$ is $1$ to $1$ but here one has to make use of the fact that determinant is invertible mod $p$, Is the following attempt correct (where $\vec{\bullet}$ means a vector and $\bar{\bullet}$ means modulo out the entries)
$\bar{\vec{0}}\neq \bar{\vec x}=\overline {I_n\cdot \vec x}=\overline{ A^{-1}A}\cdot \vec x=\overline{ A^{-1}}\overline A\cdot \vec x=\overline{ A^{-1}}\left(\overline A\cdot \vec x\right)$
What if $\det A=0$ mod $p$, which step would be incorrect ?
Your matrix $A$ has integer coefficients, so $A$ is invertible if and only if its determinant is $1$ or $-1$. So your reasoning is "almost" correct : indeed, $\overline{A}$ is invertible in $\mathbb{Z}/p\mathbb{Z}$. (Because the determinant of '$A$ modulo $p$' is equal to 'the determinant of $A$' modulo p).
But in general, $A$ is not invertible in $\mathbb{Z}$ if you took, for example, $A = \pmatrix{ 2 & 6 \\ 4 & -2},x = \pmatrix{ 1\\ 0},p = 2$, you can clearly see your reasoning is wrong (because $A$ is not invertible modulo $2$).
(Sidenote : I understand some people prefer the notation $\vec{x}$ for vectors but this is really hard to read and not that useful).