I think the counterexample given in my comment works. Let $X^3$ be the Poincare homology sphere. Let $\tilde X \to X$ be the universal cover (note: $\tilde X$ is $S^3$). As in my comment, if there were a map $S^3 \to X$ inducing a homology equivalence, then that map must necessarily factor through $\tilde X$ (by covering space theory). But this is impossible since $\tilde X \to X$ has degree 120, whereas the composite $S^3 \to \tilde X \to X$ is supposed to have degree one.

As I was writing this answer, Oscar beat me to the punch. I will keep it posted anyway. 

Let $X^3$ be the Poincare homology sphere. Let $\tilde X \to X$ be the universal cover (note: $\tilde X$ is $S^3$). As in my comment, if there were a map $S^3 \to X$ inducing a homology equivalence, then that map must necessarily factor through $\tilde X$ (by covering space theory). But this is impossible since $\tilde X \to X$ has degree 120, whereas the composite $S^3 \to \tilde X \to X$ is supposed to have degree one.