Homotopy equivalence between quotients by free actions Let $X,Y$ two contractible spaces. Assume there is a free action of a group $G$ on both spaces. 
$X$ and $Y$ are obviously homotopy equivalent. In particular, we can consider the homotopy equivalence given by the projections 
$X\xleftarrow{p_X} X\times Y\xrightarrow{p_Y}Y$

Does this equivalence induce an equivalence between the quotient spaces $X/G$ and $Y/G$?

My motivation is proving that any two little disks operads are equivalent. I'm not going to define what an operad is because it is irrelevant, but we can think of a family of contractible spaces $X_n$ for $n\in\mathbb{N}$. If each $X_n$ is equipped with a free action of the pure braid group $PB_n$, then the family $X_n/PB_n$ is called a little disks operad. The way I'm trying to prove it is inspired by this paper (page 3 of the pdf).
I believe that my question can be reduced to, given two spaces $X_G$ and $Y_G$ with $\pi_1(X)=\pi_1(Y)=G$, a homotopy equivalence between the (contractible) universal coverings $\widetilde{X}_G\to \widetilde{Y}_G$ induces a homotopy equivalence $X_G\to Y_G$. 
Since, in this case, the unviersal coverings are contractible, $X_G$ and $Y_G$ are both aspherical and path connected, so we get isomorphisms between all homotopy groups, but that's not enought to say they're homotopy equivalent.
 A: No, this is not true without additional hypotheses.  For instance, let $X$ be $G$ with the indiscrete topology; then $G$ acts on $X$ freely and $X$ is contractible but $X/G$ is a point.  On the other hand, if $Y=EG$ is the universal bundle on the classifying space of $G$ then $Y/G=BG$ is not contractible if $G$ is nontrivial.
It is true if you assume that the actions are nice enough so that $X/G$ and $Y/G$ are CW-complexes and the quotient maps $X\to X/G$ and $Y\to Y/G$ are covering maps.  Indeed, in that case $X/G$ and $Y/G$ are both $K(G,1)$ spaces and so are homotopy equivalent.  Or, going through the product as you suggest, the projection $X\times Y\to X$ is $G$-equivariant (for the product action on $X\times Y$) and so induces a map $X\times Y/G \to X/G$.  This map is a fiber bundle with fiber $Y$ (it is trivialized over any open subset of $X/G$ that is evenly covered by $X\to X/G$) and thus a homotopy equivalence since $Y$ is contractible.  Similarly, the other projection gives a homotopy equivalence $X\times Y/G\to Y/G$ and combining them we get a homotopy equivalence $X/G\to Y/G$.
