Let $[X,S^3]$ be the set of homotopy classes of maps from the pointed CW complex X to the pointed 3sphere. The group structure on S^3 endows this set with a group structure. Is this group ever noncommutative?
Yes, by Yoneda's lemma. To generalize a bit, you have a group object $G$ in some category with finite products, and you ask whether the functor it represents is pointwise abelian. This is the same as asking whether two natural transformations $[,G\times G]=[,G]\times[,G]\to[,G]$ are equal (one being $(f,g)\mapsto f\cdot g$ and the other being $(f,g)\mapsto g\cdot f$). By Yoneda, this is this case iff the representing maps $G\times G\to G$ are equal. That is, the group $[X,G]$ is always abelian iff the multiplication map on $G$ is itself commutative. In the case of $G=S^3$ in the homotopy category of spaces, the group operation is not (homotopy)commutative, though I don't know an easy proof of that off the top of my head. If you want an explicit example of an $X$ that makes $[X,G]$ nonabelian, you just have to unravel the proof of Yoneda's lemma. That means you set $X=G\times G$ and consider the identity map $G\times G\to G\times G$, which corresponds the two projection maps $p_0,p_1\in[G\times G,G]$ when you identify $[G\times G,G\times G]=[G\times G,G]\times [G\times G,G]$. So in your case, the group $[S^3\times S^3, S^3]$ is nonabelian, and in particular the two projection maps do not commute. 

