Thank you very much for the references. I'll look them up and see if I have enough knowledge to understand them.

Clearly, if you have a manifold where the symplectic structure is exact (eg a cotangent bundle $(T^*L, d\lambda)$), and if $X$ is a symplectic vector field leaving $\lambda$ invariant, then $ H_X = \lambda(X) $ is a Hamiltonian function, because $dH_X = d\iota_X \lambda = L_X \lambda - \iota_X d\lambda = - \iota_X \omega$. If $G$ is a compact Lie group acting, average first $\lambda$ over $G$ to obtain an invariant primitive for $\omega$ so that such an action is always Hamiltonian. You will either need to consider actions of non-compact groups or symplectic manifolds that are non-exact.

@ChrisGerig I did not really think through all of the details needed to construct the sequence $\{L_k\}$, but any trivialization of $TM|_\gamma$ is isomorphic to any other one. A priori the only difference is that we have $S^1 \times \mathbb{C}^n$, with a certain section $\sigma$ given by the tangent direction of the loop. I think that you can split off $\mathbb{C}\cdot \sigma$ from $S^1 \times \mathbb{C}^n$, the remaining complementary subbundle is still trivial, and we can find then a totally real subbundle to obtain any desired Maslov index, don't we? :|

@ChrisGerig to construct some $L$, I would simply trivialize the pull-back bundle $f^* TM$ (as a complex bundle). This way I have a trivialization of $TM$ over $\gamma$. Next, I choose a totally real subbundle of $TM|_\gamma$ that contains the direction of the loop. If I apply the exponential map to this subbundle, I get a submanifold, and this submanifold is still close to $\gamma$ totally real. (maybe I need to choose a Riemannian metric such that $\gamma$ is totally geodesic or so?)

@LSpice I did not know about HSM. Maybe you are right, but I think migrating my question to HSM is just the same as deleting it.

@DeaneYang Thank you very much for this story. This is very interesting. I would have expected that somebody had come up with the conjecture, and that different people would have worked in parallel to solve the claim, but not that they also came up independently with the statement. I can imagine what a shocking moment this must be to discover somebody else is just giving a talk about "your" result.