Suppose we have some 2-dimensional non-aspherical finite CW-complex $K$ with $\pi_1(K)=G$. Is there any sufficient condition on $H\leq G$ (and maybe on the group $G$ itself) which allows to conclude that the covering complex $K_H$ corresponding to the subgroup $H$ has $H_2(K_H) = 0$?

For example, take any $G=\langle x_1,...,x_n|r_1,...,r_m\rangle$ with $\text{cd}(G)>2$ and such that the words $r_1,...,r_m$ become linearly independent in abelianized free group on generators $x_1,...,x_n$. Now, let $K$ be a standard complex associated to the presentation of $G$, then $H_2(K) = 0$. So this gives a lot of examples of such $K$ for the easiest case $H=G$.

Can we get more examples with $H$ a proper subgroup of $G$?