Do we have specific examples of h-cobordant smooth manifolds $M$ and $M'$ such that $\operatorname{BDiff}(M) \not \simeq \operatorname{BDiff}(M')$? Perhaps something can be said in terms of K-theory by analyzing bundles of h-cobordisms?
In fact, if it helps to stabilize, I would be happy to know that $\operatorname{BDiff}_{\partial}(M \times I) \not \simeq \operatorname{BDiff}_{\partial}(M’ \times I)$.
This is regarding your second question.
In dimensions $\geq 5$, where the $s$-cobordism theorem applies, $h$-cobordisms are invertible in the following sense: if $W : M \leadsto M'$ is an $h$-cobordism, then there exists an $h$-cobordism $W' : M' \leadsto M$ such that $W' \circ W \cong M \times [0,1]$, and $W \circ W' \cong M' \times [0,1]$. In other words, $W$ embeds into $M \times [0,1]$ relative to $M \times \{0\}$ and so on. This can be used to obtain maps $$B\mathrm{Diff}_\partial(M \times [0,1]) \overset{W \circ -}\to B\mathrm{Diff}_\partial(W) \overset{W' \circ -}\to B\mathrm{Diff}_\partial(W' \circ W) \cong B\mathrm{Diff}_\partial(M \times [0,1])$$ which are homotopy inverses to each other. But similarly $$B\mathrm{Diff}_\partial(M' \times [0,1]) \overset{- \circ W}\to B\mathrm{Diff}_\partial(W) \overset{- \circ W'}\to B\mathrm{Diff}_\partial(W \circ W') \cong B\mathrm{Diff}_\partial(M' \times [0,1])$$ are homotopy inverses, and so $$B\mathrm{Diff}_\partial(M \times [0,1]) \simeq B\mathrm{Diff}_\partial(W) \simeq B\mathrm{Diff}_\partial(M' \times [0,1]).$$
In the comments @archipelago suggests that one should think about the case of block diffeomorphisms. Consider the (semi-simplicial) group $\widetilde{\mathrm{Diff}}(W)$ of block diffeomorphisms of $W$ which do not fix the boundaries pointwise but preserve each of the two boundary components setwise. There are maps $$B\widetilde{\mathrm{Diff}}(M) \leftarrow B\widetilde{\mathrm{Diff}}(W) \to B\widetilde{\mathrm{Diff}}(M') \tag{1}$$ given by restriction, whose fibres are $B\widetilde{\mathrm{Diff}}_M(W)$ and $B\widetilde{\mathrm{Diff}}_{M'}(W)$. Now by the same kind of reasoning as above (namely gluing on $W'$) we have $$B\widetilde{\mathrm{Diff}}_M(W) \simeq B\widetilde{\mathrm{Diff}}_{M'}(M' \times [0,1])$$ but this is just the (classifying space of the) space of block concordances of $M'$, and spaces of block concordances are contractible by a kind of Alexander trick. Thus the two maps in (1) are equivalences, so indeed $$B\widetilde{\mathrm{Diff}}_\partial(M) \simeq B\widetilde{\mathrm{Diff}}_\partial(M').$$
(I don't know the answer to your original question about $B{\mathrm{Diff}}_\partial(M) \overset{?}\simeq B{\mathrm{Diff}}_\partial(M')$, but it seems quite interesting and the above suggests also considering the related question about concordances of $M$ versus those of $M'$.)
EDIT: My argument in the case of block diffeomorphisms is slightly fallacious. The fibres of the maps in (1) need not be connected, as diffeomorphisms of $M$ (or $M'$) need not extend over $W$. However, it is true that each path-component of these fibres is contractible, as I said, because it can be identified with a space of block-concordances. The conclusion of the argument is thus that $B\widetilde{\mathrm{Diff}}(M)$ and $B\widetilde{\mathrm{Diff}}(M')$ have a common covering space, so e.g. have equivalent universal covers. Remarkably, their fundamental groups can be different: this has been proved by Samuel Muñoz Echániz, and will appear in his forthcoming PhD thesis.
