My apologies for updating this very old question.
As already mentioned, the 31st homotopy group of $S^2$ is the same as the 31st homotopy group of $S^3$. Serre's mod-C theory shows that this is a finite abelian group, and moreover that for any prime $p$ the first $p$-primary torsion occurs in $\pi_{2p} S^3$. This means that we don't have to check any primes $p$ for which $2p > 31$, which leaves us with the primes $p=3,5,7,11,13$.
The main tool for calculating these is the EHP spectral sequence. For $S^3$ the EHP sequence is pretty simple: there is a long exact sequence $$ \dots \to \pi_{n-1}(S^{2p-1}) \to \pi_{n}(S^3) \to \pi_{n}(S^{2p+1}) \to \pi_{n-2}(S^{2p-1}) \to \dots $$ that goes through $n > 3$.
The next point is the stable range. At odd primes and for odd spheres, the stable range is larger than that given by the Freudenthal theorem: if $n$ is odd, the stabilization map $\pi_{n+k}(S^n) \to \pi_k^S(S^0)$ is an epimorphism if $k=(n+1)(p-1) - 2$ and an isomorphism if $k < (n+1)(p-1)-2$. This means that all the groups relevant to computing $\pi_{31}(S^3)$ appear in the stable ranges of $\pi_*(S^{2p \pm 1})$ for $p \geq 5$, and those stable ranges are moreover very sparse (they only include the "image of J", which is well-known). If I have calculated correctly, $\pi_{31}(S^3)$ includes no $p$-torsion for $p \geq 5$.
That leaves only $p=3$, where we have to switch to a different algorithm ("I cannot do this, so I better look up what Toda calculated"). Toda showed that the 3-primary part of $\pi_{31}(S^3)$ is $\Bbb Z/3$. I found this in his 2003 text, "Unstable 3-primary Homotopy Groups of Spheres", but that was simply because I had it handier.
Combining this with the information from Mark Grant's answer gives $\Bbb Z/3 \oplus \Bbb Z/2 \oplus \Bbb Z/2 \oplus \Bbb Z/2 \oplus \Bbb Z/2$.