Suppose $(M,\omega)$ is a compact symplectic manifold and $C$ a closed curve in it. Is there a Lagrangian submanifold containing $C$? I have a sequence of $J_i$holomorphic maps from a disk to $M$, and all the maps are identical on the boundary of the disk. So, if I know that the image of the boundary is contained in a Lagrangian submanifold, I can apply Gromov convergence.

Yes, such a Lagrangian submanifold does exist. The isotropic neighborhood theorem (see for instance p. 24 of Weinstein's Lectures on Symplectic Manifolds), together with the fact that all symplectic vector bundles over the circle are trivial, implies that any closed curve in a symplectic $2n$manifold has a neighborhood symplectomorphic to the standard product $(S^1\times(\epsilon,\epsilon))\times B^{2n2}(r)$, via a symplectomorphism mapping the curve to $S^1\times \{0\}\times\{0\}$. Here $B^{2n2}(r)$ denotes the standard symplectic ball of dimension $2n2$ and some small radius $r$. Now let $T$ be a Lagrangian torus in $B^{2n2}(r)$ which contains the origin (for instance you could use a product of smallradiuscirles containing the origin in each of the $n1$ factors of $\mathbb{R}^2$ in $\mathbb{R}^{2n2}$). Then the image of $S^1\times 0\times T$ under the Weinstein symplectomorphism will be a Lagrangian submanifold containing your curve. 

