Homology equivalence may be defined as follows (any other ways could be not equivalent to one below): $X \sim Y$ if there exist two map $f : Y \to X$, $g : X \to Y$, such that $(fg)^* = id_{H(X)}$ and $(gf)^* = id_{H(Y)}$ (where homology is taken with integer coefficients).
For every CW-complex $X$, does there exist a group $\pi$ such that the Eilenberg–MacLane space $K(\pi, n)$ is homology equivalent to X?
No. For instance, take $X=S^2$ and suppose $f:X\to K(\pi,n)$ is a homology equivalence. Since $f$ is non-nullhomotopic, $\pi_2(K(\pi,n))\neq 0$, so $n=2$. But then $X$ and $K(\pi,n)$ are simply connected, so $f$ would have to be a homotopy equivalence. Since $X$ has higher homotopy groups, this is a contradiction.
More generally, a similar argument works whenever $X$ is simply connected and not an Eilenberg-MacLane space (by the Hurewicz theorem, a homology equivalance $f:X\to K(\pi,n)$ must induce a nontrivial map on the first nontrivial homotopy group of $X$, and it follows that $n>1$ so $f$ must be a homotopy equivalence).
I would note, however, that there is a positive result if you require the homology equivalence only to exist as a map in one direction. The Kan-Thurston theorem says that if $X$ is any path-connected space, then there is a group $\pi$ and a map $f:K(\pi,1)\to X$ that induces isomorphisms on homology.