I think the Reeb foliation can be generated by two independent global vector fields so the Frobenius rank of $S^{3}$ is "2".
The reason is that every real vector bundle on $S^{3}$ is a trivial bundle. so the two dimensional subvector bundle of $TS^{3}$ tangent to the Reeb foliation, is a trivial bundle. So there are two global continuous sections for this distribution. (These two sections can be choosed smooth by standard approximations). This proves our claim.
The reason that every n-dim real bundle on $S^{3}$ is trivial:
Using Technics of clutching functions, explained in "K-theory and vector bundles" by Allen Hatcher we conclude that the equivalent classes of n dim vector bundles on $S^{k}$ is isomorphic the homotopy class of continuous functions from $S^{k-1}$ to $GL(n,\mathbb{R})$. In our particular case $S^{3},\;\;k=3,\;n=2$, we have that $\pi_{2}(GL(2,\mathbb{R}))$ is trivial, see Homotopy groups of Lie groups So the only n. dim real bundle on $S^{3}$ is trivial.
So why it is not customary to introduce the Reeb foliation by two explicit global vector fields tangent to $S^{3}$? On the other hand, it is well known that the Reeb foliation can not be generated by two Analytic vector fields. So it is natural to ask:
Is there a global analytic vector field tangent to the Reeb foliation?(We can consider both singular and non singular case. In the singular case, what would be the nature of singularities?)
