Yes for "hope" 1. This theorem was proven by Moser using volume-preserving flows. A manifold with a volume form is the same thing as a manifold with an atlas of charts modeled on the volume-preserving diffeomorphism pseudogroup. He found an argument that can be adapted to either the symplectic case or the volume case. I cited this result in my paper A volume-preserving counterexample to the Seifert conjecture (Comment. Math. Helv. 71 (1996), no. 1, 70-97), where I also established a similar result for the volume-preserving PL pseudogroup. In the PL case, the corresponding decoration on the manifold is a piecewise constant volume form.
In my opinion, the most exciting result on this theme is the Ulam-Oxtoby theorem. (But Moser's version is the most useful one and the most elegant.) The theorem is that if you have a topological manifold with any Borel measure that has no atoms and no bald spots, then it is modeled on the pseudogroup of volume-preseserving homeomorphisms. For example, you can start with Lebesgue measure in the plane and add uniform measure on a circle, and there is a homeomorphism that takes that measure to Lebesgue measure.
For a long time I have wondered about the pseudogroup of volume-preserving Lipschitz maps. The question is whether there is a corresponding cone of measures, and if so, how to characterize it.
My reading of the question is this: we're given $H\in C^\infty(M)$ with $M$ symplectic, and we want to know whether there's a submanifold $L\subset M$, a Riemannian metric $g$ on $L$, and a symplectomorphism $T^\ast L \cong M$ under which $H$ pulls back to the norm-square function. And we want to know if $(L,g)$ is unique.
Uniqueness is easy: we recover $L$ as $H^{-1}(0)$, and $g$ as the Hessian form of $H$ on the vertical tangent bundle (determined by the symplectomorphism) along $L$.
Basic necessary conditions:
(1) $L:=H^{-1}(0)$ is a Lagrangian submanifold of $M$.
(2) $L$ is a non-degenerate critical manifold of $H$ of normal Morse index 0.
These conditions imply that a neighbourhood of $L$ embeds symplectically in $T^\ast L$, and also (by the Morse-Bott lemma) that $H$ is quadratic in suitable coordinates near $L$. These two sets of coordinates needn't be compatible, so let's replace (2) by something much stronger (but still intrinsic):
(3) There's a complete, conformally symplectic vector field $X$ (i.e., $\mathcal{L}_X\omega=\omega$), whose zero-set is exactly $L$, along which $H$ increases quadratically (i.e., $dH(X)=2H$).
I claim that (1) and (3) are sufficient. With these data, you can locate a point $x\in M$ in $T^\ast L$. Flow $X$ backwards in time starting at $x$ to obtain the projection to $L$; pay attention to the direction of approach to $L$ to get a tangent ray, and use the metric (i.e., the Hessian of $H$ on the fibres of projection to $L$) to convert it to a cotangent ray. Pick out a cotangent vector in this ray by examining $H(x)$. If I'm not mistaken, this will single out a symplectomorphism with the desired properties.
Best Answer
You don't need to use Weinstein's tubular neighborhood theorem to assign closed one forms on L to deformations of L. Here is a construction which makes it clear the assignment is canonical.
A smooth family of Lagrangian submanifolds is given by a pair of smooth maps $$\mathbb R \xleftarrow{t}X \xrightarrow{f} M$$ so that the map $t$ is a proper submersion and $f$ includes every fiber of $t$ as a Lagrangian submanifold of $M$.
There is a vertical cotangent bundle of $X$ which is the quotient of $T^*X$ by the pullback of one forms from $\mathbb R$. This vertical cotangent bundle should be regarded as putting together the cotangent bundles of the fibers of $t$ into a smooth vector bundle over $X$. Each differential form $\theta$ on $X$ has a well defined projection to a section $\pi\theta$ of the wedge of the vertical cotangent bundle, which is the definition of a smooth family of differential forms on the fibers of $t$. The fact that this is a family of Lagrangian submanifolds implies that $\pi(f^*\omega)=0$.
Choose any smooth vector field $\frac \partial {\partial t} $ on $X$ so that $\frac\partial{\partial t} t=1$. Then $$\pi(\iota_{\frac \partial{\partial t}} f^*\omega)$$ is a family of one forms on the fibers of $t$ which does not depend on the choice of $\frac \partial {\partial t}$. It is a family of closed one forms because $\pi$ commutes with $d$ and $$\pi L_{\frac\partial{\partial t}}f^*\omega=0$$.
This construction reverses the assignment of a deformation of L to a closed one form on L which uses the Weinstein neighborhood theorem.