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.

Let me conisder the case when the distribution of planes is of codimension 1 and explain why in this case it is enough to have $C^1$ smoothness in order to ensure the existence of the folitation. 

In the case when the distribution is of codimension 1, you can formulate Frobenius Theorem in terms of 1-forms. 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 $C^1$. Let me give a sketch of the proof that $A\wedge dA=0$ garanties existence of the foliation is A is $C^1$.

The proof is by induction

1. Consider the case $n=2$. In this case it is a standard fact of ODE, that for a $C^1$ smooth distribution of directions on the plane the integral lines are uniquelly defined.

2. Conisder the case $n=3$. We will show that the foliation exists locally near any point, say the origin $O$ of $R^3$. The 1-form A, that defines the distribution is non vanishing on one of the coordinate planes, say $(x,y)$ plane in the neighborhood of $O$. Take a $C^1$ smooth vector field in the neigborhood of $O$ that is transversal to planes $z=const$ and satisfies $A(v)=0$. Take the flow correponding to this vector field. The flow is $C^1$ smooth and moreover it preserves the distribution of planes $A=0$. Indeed, dA vanishes on the planes A=0 (by the condition of integrability), and we can apply the formula for Lie derivative $L_v(A)=d(i_v(A))+i_v(dA)=i_v(dA)$. Finally, we take the integral curve of the restriction of $A=0$ to the plane $(x,y)$ and for evey curve conisder the surface it covers unders the flow of $v$. This gives the foliation.

This reasoning can be repeated by induction.

A good refference is Arnold, Geometric methods of ordinary differential equations. I don't know if this book was transalted to English