Here's another example.
Consider $\mathbb{R}P^3$ in the model of a 3-ball with antipodal boundary points identified.
Consider the $G = \mathbb{Z}/2\mathbb{Z}$ action on $\mathbb{R}P^3$ given by the antipodal map sending $(x,y,z)$ to $-(x,y,z)$.
I claim that $\mathbb{R}P^3/G$ is homeomorphic to a cone on $\mathbb{R}P^2$. To see this, use coordinates on $C\mathbb{R}P^2$ given by $([x,y,z],t)$ where we think of $(x,y,z)\in\mathbb{R}^3$ and we're collapsing $\mathbb{R}P^2\times\{0\}$ to a point. Now,map $(x,y,z)$ in $\mathbb{R}P^3$ to $\big([x,y,z], (x^2+y^2+z^2)\big)$ (and map the origin to the cone point). This is clearly continuous away from the origin. It's not too hard to see that it's continuous at the origin as well.
It's also not hard to see that this descends to a bijective map from $\mathbb{R}P^3/G$ to $C\mathbb{R}P^2$, which is therefore a continuous bijection between compact Hausdorff spaces, so is itself a homeomorphism.
Finally, notice that $C\mathbb{R}P^2$ is not a topological manifold (with or without boundary) because of the cone point $p$. A neighborhood $U$ of the cone point $p$ has, by excision, $H_k(U, U-p)\cong H_k(C\mathbb{R}P^2, C\mathbb{R}P^2-p) = H_{k-1}(\mathbb{R}P^2)$, which means $p$ can be neither a manifold point nor a manifold-with-boundary boundary point.
You can find the construction of homology with general coefficients and the universal coefficient theorem in Hatcher's Algebraic Topology, which is available free from his website.
The answer to your third question is yes.
The answer to the second part of your first question is yes, especially in the case that we take $G$ to be a field, most often finite or $\mathbb{Q}$, or $\mathbb{R}$ in differential topology. Homology over a field is simple because $\operatorname{Tor}$ always vanishes, so you get e.g. an exact duality between homology and cohomology. Homology with $\mathbb{Z}_2$ coefficients is also the appropriate theory for many questions about non-orientable manifolds-their top $\Bbb{Z}$-homology is zero, but their top $\Bbb{Z}_2$ homology is $\Bbb{Z}_2$, which leads to the degree theory in Milnor you were mentioning.
Cohomology with more general coefficients than $\mathbb{Z}$ is even more useful than homology. For instance it leads to the result that if a manifold $M$ has any Betti number $b_i(M)<b_i(N)$, where $b_i$ is the rank of the free part of $H_i$, there's no map $M\to N$ of non-zero degree. This has lots of quick corollaries-for instance, there's no surjection of $S^n$ onto any $n$-manifold with nontrivial lower homology! Edit: This is obviously false, and I no longer have any idea whether I meant anything true.
But in the end $H_*(X;G)$ is more of a stepping stone than anything else; it gets you thinking about how much variety there could be in theories satisfying the axioms of homology. It turns out there's almost none-singular homology with coefficients in $G$ is the only example-but if we rid ourselves of the "dimension axiom"
$$H_*(\star)=\left\{\begin{matrix}\mathbb{Z},*=0\\0,*>0\end{matrix}\right.$$
then we get a vast collection of "generalized (co)homology theories," beginning with K-theory, cobordism, and stable homotopy, which really do contain new information. In some cases, so much new information that we can't actually compute them yet!
Best Answer
Since you are asking about finite perfect groups, here is an example. Let $G=A_5$. Then $H_2(G)\cong {\mathbb Z}_2\ne 0$ (see here: Schur multiplier is another name for $H_2$) This group acts freely on $$ X= S^5\times S^7, $$ see Theorem 1.1 of
Fixity and free group actions on products of spheres, by A.Adem, J.Davis and O.Unlu, Commentarii Mathematici Helvetici, 2004, Vol. 79, pp 758--778.
It follows that $H_2(X/G)\cong H_2(G)\ne 0$, while $H_2(X)=0$.
If you want $A$ a proper subset of $X$, take $X$ to be the product $$ S^5\times S^7 \times S^{2019} $$ and let $G$ act trivially on $S^{2019}$. Then take $A$ to be the product $$ S^5\times S^7 \times\{p\}\subset X. $$
On the other hand, there are, of course, examples where $q(A)$ is acyclic. For instance, let $A$ be a point. Or let $X$ be a smooth manifold, $G\times X\to X$ a smooth action, $A$ a small $G$-invariant ball containing a fixed point of the $G$-action on $X$.