The answer to the title question, for the usual meaning of "homology theory," is no. Homology is always an invariant of the stable homotopy type of a space, and so no homology theory can distinguish two spaces which are stable homotopy equivalent but not homotopy equivalent. For example, no homology theory can distinguish $S^1 \times S^1$ and $S^1 \vee S^1 \vee S^2$, but the former has nontrivial cup products while the latter does not.

On the other hand, we have the following homology version of Whitehead's theorem, which bypasses the above constraint because it does not come from a homology theory in the sense that it does not even come from applying a functor to spaces.

Theorem: Suppose $f : X \to Y$ is a map of spaces inducing an isomorphism on $\pi_1$ and an isomorphism on homology with all local coefficients. Then $f$ is a weak equivalence. (In particular, if $X, Y$ are simply connected, we only need $f$ to induce an isomorphism on homology.)

The answer to the title question, for the usual meaning of "homology theory," is no. Homology is always an invariant of the stable homotopy type of a space, and so no homology theory can distinguish two spaces which are stable homotopy equivalent but not homotopy equivalent. For example, no homology theory can distinguish $S^1 \times S^1$ and $S^1 \vee S^1 \vee S^2$, but the former has nontrivial cup products while the latter does not.

On the other hand, we have the following homology version of Whitehead's theorem, which bypasses the above constraint because it does not come from a homology theory in the usual sense.

Theorem: Suppose $f : X \to Y$ is a map of path-connected spaces inducing an isomorphism on $\pi_1$ and an isomorphism on homology with all local coefficients. Then $f$ is a weak equivalence. (In particular, if $X, Y$ are simply connected, we only need $f$ to induce an isomorphism on homology.)