### preliminary remark

I assume that being independant for functions here means that the their differentials at **any** point are linearly indenpendant (and not at **almost**, like it is sometimes assumed... in which case a counter-example is easy to find for $n=2$: take $f_1$ the square of the norm on $\mathbb{R}^2$, and look at the origin.  

### the statement

I clam that one can prove the following version of the action-angle coordinate Theorem, under the hypothesis of the question: 

> For any $x\in M$ there exists Darboux coordinates $(p_1,\dots,p_n,q_1,\dots, q_n)$  
> around $x$ such that the leaves of the Lagrangian foliation are $(q_1,\dots,q_n)=cst$. 

Then the local manifold $\{p_1=\dots=p_n=0\}$ satisfies your requirement. 

To prove the statement, consider $q_i=f_i-f_i(x)$ and extend them to a Darboux chart $(q_1,\dots,q_n,p_1,\dots,p_n)$ around $x$. 

Tell me if I have misunderstood the question.