Is every contractible loop contained in a Darboux chart?

Let $(M,\omega)$ be a symplectic manifold and $\gamma:S^1 \rightarrow M$ be a contractible smooth loop. Is it possible to find an open set $U \subset M$ such that $\gamma(S^1) \subset U$ and such that there exists a Darboux chart $\phi : U \rightarrow \mathbb{R}^{2n}$?

Clearly this isn't true if $\gamma$ is not assumed contractible (there might not be any chart on $M$ that contains $\gamma(S^1)$!). If $\gamma$ is contractible then there certainly do exists charts that contain $\gamma(S^1)$, but do there necessarily exist Darboux charts?

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Yes, it is always possible. Of course, if $M$ has dimension $2$ we need to assume that $\gamma$ is embedded in $M$ in which case the statement is easy (but if $\gamma$ have self-intersections, the statement does not hold necessarily). If $dim(M)$ is at least $4$, then the statement can be deduced from the following two:
1) Let $\gamma_t$ be a smooth family of embedded loops in $M$ such that for any $t_1,t_2$ the integral of $\omega$ over the cylinder swiped by loops $\gamma_t$, $t\in [t_1,t_2]$ is zero. Then for any $t_1,t_2$ there is a Hamiltonian isotopy on $M$ that takes $\gamma_{t_1}$ to $\gamma_{t_2}$. This can be proven by constructing a family of Hamiltoninas $H_t$ parametrised by $t$ and considering the corresponding flow on $M$ (infinitesimally $H_t$ takes $\gamma_t$ to $\gamma_{t+\Delta t}$).
2) Provided $dim(M)>2$ each contractible loop can be contracted to an arbitrarily small neighbourhood of a point on $M$ by the homotopy having the property of 1) (i.e. preserving symplectic action).