It won’t exist in general (eg if $Aut(\Gamma)$ is simple.

Yes, the Borel $\sigma$-algebra.

This should work in dimension 2 also? (So Ghomi’s comment on the monkey saddle doesn’t seem correct)

The centralizer of a hyperelliptic, quotient the hyperelliptic, is a mapping class group of a $2g+2$ pointed sphere. So essentially a braid group.

One natural way such a homomorphism could exist is if there is a section of the homomorphism of the handlebody group to $Out(F_g)$. But such a section does not exist for $g>3$ (and it exists for $g=2$ as observed in my answer below). mathematik.uni-muenchen.de/~hensel/papers/nosplit2.pdf

For complex hyperbolic manifolds (compact ball quotients), signature is $1/3$ of Euler characteristic. en.m.wikipedia.org/wiki/… and is proportional to volume. Complex hyperbolicity is also a sufficient condition to say that the fundamental group is a free product and any homemorphism must respect this free product, so preserves the two factors (up to conjugacy of each factor).

I’m guessing one could arrange signature 0 non-aspherical by taking a connected sum of manifolds with opposite signatures and which are not related by a reflection. I think a connected sum of non-isometric complex hyperbolic manifolds of the same volume (but one with orientation reversed ) should work.

I’m pretty sure that one could find Gromov-Piatetskii-Shapiro examples of hyperbolic 4-manifolds which have this property. Arithmetic hyperbolic 4-manifolds are commensurable with non-orientable orbifolds. The GPS examples are obtained by “interbreeding” incommensurable hyperbolic manifolds by cutting and pasting along isometric totally geodesic hypersurfaces. By making sure that non-orientable symmetries don’t agree along the gluing, I think one could arrange that they do not cover a non-orientable orbifold. But of course this would require a careful construction.