Certainly. There's a general description of all compact 3-manifolds now that geometrization is about.
So for homology 3-sphere's you have the essentially unique connect sum decomposition into primes.
A prime homology 3-sphere has unique splice decomposition (Larry Siebenmann's terminology). The splice decomposition is just a convenient way of encoding the JSJ-decomposition. The tori of the JSJ-decomposition cut the manifold into components that are atoroidal, so you form a graph corresponding to these components (as vertices) and the tori as edges.
The splice decomposition you can think of as tree where the vertices are decorated by pairs (M,L) where M is a homology 3-sphere and L is a link in M such that M \ L is an atoroidal manifold.
By geometrization there's not many candidates for pairs (M,L). The seifert-fibred homology spheres that come up this way are the Brieskorn spheres, in that case L will be a collection of fibres in the Seifert fibering. Or the pair (M,L) could be a hyperbolic link in a homology sphere. That's a pretty big class of manifolds for which there aren't quite as compact a description, compared to, say, Brieskorn spheres.
I wrote a little expository piece about this and related matters in the Newsletter of the European Mathematical Society:
http://www.ems-ph.org/journals/newsletter/pdf/2010-03-75.pdf
The classical topology of $X:=T^\ast L$ can be taken to include a little more than its diffeomorphism type: there's also an almost complex structure $J$ on $X$, canonical up to homotopy. The pair $(X,J)$ knows the Pontryagin classes of $L$, because $$c_{2k}(TX,J)|_{L}=c_{2k}(TL\otimes\mathbb{C})=(-1)^k p_k(TL),$$
so $(-1)^k p_k(TL)$ pulls back to $c_{2k}(L)$ under projection $X\to L$. However, even with this embellishment, the smooth topology of $X$ doesn't determine $L$.
Faithfulness conjecture: the exact symplectomorphism type of the cotangent bundle $(X,\omega=d\lambda_L)$ of a compact manifold $L$ determines $L$.
An exact symplectomorphism $T^\ast L \to T^\ast L'$ is a diffeomorphism $f$ such that $f^*\lambda_{L'}-\lambda_L= dh$ for $h$ a compactly supported function. The conjecture (but not the name) is standard.
Attempts to use symplectic invariants of $X$ to distinguish smooth structures on $L$ have so far been a complete failure. For example, the symplectic cohomology ring $SH^*(X)$ is isomorphic to loopspace homology $H_{-*}(\mathcal{L}L; w)$ (the coefficients are the local system $w$ of $\mathbb{Z}$-modules determined by $w_2$), with the string product. This invariant is determined by the homotopy type of $L$.
"Arnol'd's conjecture" (scare quotes because Arnol'd really made a much more circumspect conjecture). Any exact Lagrangian embedding $\Lambda \to X$ (with $\Lambda$ compact) is exact-isotopic to the embedding of the zero-section.
This would immediately imply the faithfulness conjecture.
There has been progress towards Arnol'd's conjecture of three kinds:
(1) It's true for $L=S^2$ (Hind, http://arxiv.org/abs/math/0311092).
(2) The work of several authors (Fukaya-Seidel-Smith http://arxiv.org/abs/0705.3450, Nadler http://arxiv.org/abs/math/0612399, Abouzaid http://arxiv.org/abs/1005.0358, and Kragh's work in progress) cumulatively shows that the projection from an exact Lagrangian to the zero-section is a homotopy equivalence. This is good evidence for the truth of the conjecture, but for the application to faithfulness one might as well make homotopy-equivalence an assumption.
(3) As Andy mentioned, Abouzaid http://arxiv.org/abs/0812.4781 has shown that a homotopy $(4n+1)$-sphere $S$, such that $T^*S$ contains an exact embedded Lagrangian $S^{4n+1}$, bounds a parallelizable manifold. This is proved by a stunning analysis of the geometry of a space of pseudo-holomorphic discs.
The existence of exact Lagrangian immersions is governed by homotopy theory (there is an h-principle which finds such an immersion given suitable homotopical data). Just as the subtleties of 4-manifold topology can be located at the impossibility of removing double points of immersed surfaces (the failure of the Whitney trick), so the subtlety of the symplectic structure of cotangent bundles comes down to the question of removability of double points of Lagrangian immersions.
Best Answer
Yes, there's loads of other contractible 4-manifolds bounding homology spheres with other geometries.
For example $1$-surgery on the Stevedore knot is a hyperbolic manifold with volume $1.3985\cdots$. This bounds a contractible manifold by the same kind of arguments Casson and Harer used to cook up their big list of Mazur manifolds. There are other examples like this that occur in my arXiv preprint: https://arxiv.org/abs/0810.2346 Some are geometric, some have incompressible tori.
I don't think any Nil manifolds bound contractible 4-manifolds. Crisp and Hillman determined all the Nil manifolds that embed smoothly (or topologically) in the $4$-sphere. The only Nil manifolds on that list have non-trivial homology. Crisp & Hillman "Embedding Seifert fibred 3-manifolds and Sol-manifolds in 4-space".