This is just a remak, no the answer. In the case the distribution is of codimension 1, you can reformulate Frobenius Theorem, so that it uses just one derivative. Namely you can define a non-zero 1-form $A$, whose kernel is the distribution. The smoothness of this 1-form will be the same as the smoothness of the distribution. Now, you can say that the distribution is integrable if $A\wedge dA=0$. This quantity is well defined is A is $C^1$. Though it is not clear for me if this garatinies the integrablility.