(This answer has been edited to give more details.)

Finitely generated homotopy groups do not imply finitely generated homology groups. Stallings gave an example of a finitely presented group $G$ such that $H_3(G;Z)$ is not finitely generated. A $K(G,1)$ space then has finitely generated homotopy groups but not finitely generated homology groups. Stallings' paper appeared in Amer. J. Math 83 (1963), 541-543. Footnote in small print: Stallings' example can also be found in my algebraic topology book, pp.423-426, as part of a more general family of examples due to Bestvina and Brady.

As Stallings noted, it follows that any finite complex $K$ with $\pi_1(K)=G$ has $\pi_2(K)$ nonfinitely generated, even as a module over $\pi_1(K)$. This is in contrast to the example of $S^1 \vee S^2$.

Finite homotopy groups do imply finite homology groups, however. In the simply-connected case this is a consequence of Serre's mod C theory, but for the nonsimply-connected case I don't know a reference in the literature. I asked about this on Don Davis' algebraic topology listserv in 2001 and got answers from Bill Browder and Tom Goodwillie. Here's the link to their answers:

http://www.lehigh.edu/~dmd1/tg39.txt

The argument goes as follows. First consider the special case that the given space $X$ is $BG$ for a finite group $G$. The standard model for $BG$ has finite skeleta when $G$ is finite so the homology is finitely generated. A standard transfer argument using the contractible universal cover shows that the homology is annihilated by $|G|$, so it must then be finite.

For a general $X$ with finite homotopy groups one uses the fibration $E \to X \to BG$ where $G=\pi_1(X)$ and $E$ is the universal cover of $X$. The Serre spectral sequence for this fibration has $E^2_{pq}=H_p(BG;H_q(E))$ where the coefficients may be twisted, so a little care is needed. From the simply-connected case we know that $H_q(E)$ is finite for $q>0$. Since $BG$ has finite skeleta this implies $E_{pq}^2$ is finite for $q>0$, even with twisted coefficients. To see this one could for example go back to the $E^1$ page where $E_{pq}^1=C_p(BG;H_q(E))$, the cellular chain group, a finite abelian group when $q>0$, which implies finiteness of $E_{pq}^2$ for $q>0$. When $q=0$ we have $E_{p0}^2=H_p(BG;Z)$ with untwisted coefficients, so this is finite for $p>0$ by the earlier special case. Now we have $E_{pq}^2$ finite for $p+q>0$, so the same must be true for $E^\infty$ and hence $H_n(X)$ is finite for $n>0$.

Sorry for the length of this answer and for the multiple edits, but it seemed worthwhile to get this argument on record.