Assume spaces to be connected. In general, the set of unbased homotopy classes $[X,Y]$ differs from the set of pointed homotopy classes $\langle X,Y\rangle$. This is captured by the $\pi_1(Y)$ action on the set $\langle X,Y\rangle$. The (computable) examples I know for which this action is non-trivial are not manifolds. What's the story with manifolds?
Let's do a simple example. Suppose we consider a connected manifold $M$, together with a chosen basepoint $b$. Consider $m\in M$ away from the basepoint and unbased homotopy classes of loops, for which a representative loop runs through $m$. Can we always pick a whisker from $b$ to $m$ to get based homotopy classes of loops (with basepoint $b$) such that forgetting the basepoint gives back the same homotopy class?
In my mind, this is a question about choosing a whisker that does not pick up any $\pi_1$-action. I'm not entirely sure, but I think that if the target space is connected, there is a surjection $\langle X,Y\rangle\rightarrow [X,Y]$. Is the right-inverse choosing a suitable whisker?
If $\gamma\colon S^1\rightarrow M$ is some unbased loop and $m=\gamma(1)$, then picking a path $\eta$ from the basepoint $b$ to $m$ in $M$ (here, we use that $M$ is connected) yields a based loop $\eta\gamma\eta^{-1}$ in $M$ and this loop is freely homotopic to $\gamma$ (the homotopy traverses at time $t$ the loop $\eta\vert_{[t,1]}\gamma\eta\vert_{[t,1]}^{-1}$). If $\eta^{\prime}$ is another such path from $b$ to $m$, then $\eta^{\prime}\eta^{-1}$ is a based loop in $M$ and it operates taking $\eta\gamma\eta^{-1}$ to $\eta^{\prime}\gamma{\eta^{\prime}}^{-1}$. Thus, distinct choices of such "whiskers" realize distinct elements in the same orbit of the $\pi_1(M,b)$-action (and, conversely, one can show that any two elements in the same orbit can be transformed into another by such a choice). In particular, this argument demonstrates that we have a surjection $\pi_1(M,b)=\langle S^1,M\rangle\twoheadrightarrow[S^1,M]$.
The same is true if we replace $S^1$ by any well-pointed space $X$ with basepoint $x_0$. In this case, "attaching a whisker" means taking the space $X^{\prime}=X\vee I$, where $x_0$ is glued to $0$ and we take $1$ as the basepoint of $X^{\prime}$. The inclusion $i\colon X\rightarrow X^{\prime}$ and projection $r\colon X^{\prime}\rightarrow X$ are easily seen to be inverse homotopy-equivalences. Note that $i$ is unbased, but $r$ is based. However, $X$ being well-pointed implies that there exists a based map $j\colon X\rightarrow X^{\prime}$ s.t. $r$ and $j$ are inverse pointed homotopy-equivalences. Now, taking an unbased map $f\colon X\rightarrow M$, we construct a based map $F\colon X^{\prime}\rightarrow M$ s.t. $F\vert_X=f$ and $F\vert_I$ is a path from $f(x_0)$ to $b$. Then, $Fj\colon X\rightarrow M$ is a based map that is freely homotopic to $f$ since there are unbased homotopy-equivalences $Fj\simeq Firj\simeq Fi=f$. Thus, we have a surjection $\langle X,M\rangle\twoheadrightarrow[X,M]$. Nothing here relies on $M$ being a manifold, for the record.
To answer your other question, any manifold with non-abelian fundamental group is, in particular, a non-simple manifold. For example, any compact, connected surface except for $T^2$, $S^2$ or $\mathbb{RP}^2$ yields an example. Note that these all are connected sums. In higher dimensions, this becomes even easier: If $M,N$ are manifolds of the same dimension $\ge3$, then their connected sum $M\#N$ has fundamental group $\pi_1(M)\ast\pi_1(N)$, hence is non-abelian if neither $M$ nor $N$ is simply connected. To generalize in another direction, whilst $\mathbb{RP}^2$ has abelian fundamental group, it is still not a simple space, because $\pi_1(\mathbb{RP}^2)=\mathbb{Z}/2$ acts on $\pi_2(\mathbb{RP}^2)=\mathbb{Z}$ by multiplication with $-1$. This generalizes to all other even-dimensional projective spaces, and even to Grassmannians $\mathrm{Gr}(k,n)$, where $k<n$ is even.