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 $S^{3},\;\;k=3$ we have  that $\pi_{2}(GL(2,\mathbb{R})$ is  trivial, see http://mathoverflow.net/questions/8957/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 $S^{3}$?