(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.
Autumn's answer captures the essence why there is a $\mathbb{Z}_2$ is in the first homology
of a closed nonorientable surface.
If you remove a disk from a closed surface, the resulting object has a $1$-dimensional $CW$-complex as a strong deformation retract, so that the homology of the resulting object
has no torsion.
A closed nonorientable surface is always the result of the connect sum of an orientable surface
with a projective plane or two projective planes. That is you are gluing one Moebius band, or two Moebius bands into the boundary of an orientable surface. The core or cores of the Moebius bands, oriented appropriately represent the generator of the $2$-torsion in the first homology of the surface.
Best Answer
Homology also has complicated and unintuitive aspects if one goes beyond nice spaces like CW complexes. A surprising example of this is the subspace of Euclidean 3-space consisting of the union of a countable number of 2-spheres with a single point in common and the diameters of the spheres approaching zero. (This is the 2-dimensional analog of the familiar "Hawaiian earring" space.) Then the amazing fact is that the n-th homology group of this space with rational coefficients is nonzero, and even uncountable, for each n > 1. This was shown by Barratt and Milnor in a 1962 paper in the AMS Proceedings.
The result holds also with integer coefficients, with homology classes that are in the image of the Hurewicz homomorphism. So one could say that this strange behavior comes from homotopy groups but just happens to persist in homology. I would guess that there are also examples where the homology is not in the image of the Hurewicz homomorphism, so it does not come directly from homotopy groups.