(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.
In Donald Kahn's 1966 paper "On stable homotopy modules" (Inventiones 1, pp 375–379, doi:10.1007/BF01389739)
he constructs examples of CW complexes with 3 cells, whose stable homotopy groups are not finitely generated as modules over the ring of stable homotopy groups of spheres. I believe this answers in the negative a stable analogue of the question.
Perhaps a Freudenthal suspension theorem argument, together with the stable result, can imply the desired unstable result. I am sorry but I do not have a few minutes to spare to explore this possibility; perhaps others are interested enough to try it out, though. (I am sorry if this incomplete answer is inappropriate for this site, which I am a beginner to. Please feel free to delete this answer if so. I would certainly have made this all a comment rather than an answer, but don't have enough points on this site to leave comments.)
Best Answer
By results of J.P. Serre (for $p=2$) and Y. Umeda (for odd $p$) we know that a 1-connected finite CW-complex $X$ with non-trivial cohomology mod $p$ has infinitely many non-trivial homotopy groups mod $p$.
In fact C.A. McGibbon and J.A. Neisendorfer have proved the existence of $p$-torsion elements in infinitely many dimensions for finite dimensional spaces using H. Miller's theorem on the Sullivan conjecture.