Homotopy of space of immersions, Smale-Hirsch theorem

No. For example, if $M$ is a Moebius band then, at least for even $k$, $Imm(M,\mathbb R^{2+k})$ is not homotopy equivalent to $Imm(S^1\times \mathbb R,\mathbb R^{2+k})$.

The latter is equivalent to the space of all maps from $S^1$ to $O(2+k)/O(k)$. Its first non-trivial homotopy group is $\pi_{k-1}=\pi_k(O(2+k)/O(k))=\mathbb Z$.

The former is equivalent to the space of sections of a nontrivial bundle with base $S^1$ and fiber $O(2+k)/O(k)$, in which the action of the fundamental group of the base on $\pi_k(O(2+k)/O(k))$ is nontrivial. Its first non-trivial homotopy group is $\mathbb Z/2\mathbb Z$.

Note that when $M$ and $M′$ are simply connected, I have a positive answer in the rational case. I.e

If $k\geq m$ is an odd integer, then $M\simeq M'\Rightarrow Imm\left(M,\mathbb{R}^{m+k}\right)\simeq_{\mathbb{Q}}Imm\left(M',\mathbb{R}^{m+k}\right).$

Here $"\simeq_{\mathbb{Q}}"$ denotes the rational homotopy equivalence.