Isn't it a consequence of the existence of action-angle semi-global coordinates? Namely, one has that 

> If $F=(f_1,\dots,f_n)$ is proper, then for any lagrangian invariant torus 
> $\Lambda$ there exists a neighborhood $U$ of $\Lambda$ that is symplectomorphic to  
> a neighborhood $V$ of a fiber of the projection in the cotangent bundle $T^*\mathbb{T}^n$ 
> of the $n$-dimensional torus $\mathbb{T}^n$ (namely, $V=\{(\theta,I), |I-I_0|\leq r\}$). 

It seems to me that the zero section $\{\theta=0\}$ satisfies your requirement. 

EDIT: if you read french then you can find a proof of this statement in [Lecture Notes by Colin de Verdière][1] (proof of Theorem 20, pages 50-51). 

EDIT 2: I think the proof is the same as the one in Colin de Verdiere's notes. Namely, you first prove it in the dimension $1$ situation. Here you don't need properness assumption since the conclusion you are asking for is only a local statement and not a semi-global one (it is about the neighborhood of a point, while action-angle is about the neighborhood of a Lagrangian leaf). The rest of the proof is the same except that instead of a free action of $\mathbb{T}^n$ you have a locally free action of $\mathbb{R}^n$ (which is the only thing you need for your conclusion). 

Obvious remark concerning vocabulary: there is one more thing to care about. It is what people mean by $f_1,\dots,f_n$ being independant. I personnaly mean that at **almost** any point their derivatives are linearly indpendant. Therefore in this case the conclusion is true for **almost** all points in $M$. At critical points of $F$ this is very easy to construct a counter-example in dimension $2$: take $F$ to be the square of the norm in $\mathbb{R}^2$, and look at what happens at the origin. 

  [1]: http://www-fourier.ujf-grenoble.fr/~ycolver/Livresc/intro.ps.gz