Question: Let $X$ and $Y$ be path-connected spaces that admit a contractible universal cover, with $\pi_1(X) \cong \pi_1(Y)$. Is $X$ homotopy equivalent to $Y$?
Comments: $X$ and $Y$ are both $K(\pi_1(X),1)$s. In particular, this implies that every homomorphism $\varphi: \pi_1(X) \rightarrow \pi_1(Y)$ (e.g., the isomorphism) is induced by a map $f: X \rightarrow Y$. If $X$ and $Y$ are both CW-complexes, Whitehead's theorem says that $f$ is a homotopy equivalence. (In general, by definition, $f$ is a weak homotopy equivalence.) So a counterexample requires that at least one of $X$ and $Y$ fails to have homotopy type a CW-complex.
If one removes the requirement that $X$ and $Y$ have contractible universal cover, in particular if one even relaxes it to $X$ and $Y$ have weakly contractible universal cover, the double comb space is a simply-connected counterexample, as it is not contractible. (A proof that it is not contractible can be found here.)
As proposed by studiosus in the comments, the standard unit circle and the pseudocircle (http://en.wikipedia.org/wiki/Pseudocircle) serve as a counterexample, since their universal covering spaces are the real line and the Khalimsky line, both of them contractible. (The contractibility of the latter follows from http://arxiv.org/pdf/0901.2621.pdf .)