I don't know if you will be satisfied with my answer, but in my understanding it is an extremely rare situation that a chosen loop is the boundary of a holomorphic curve.  While it might be tricky and situation specific to give you a complete and general proof of "generic" non-existence, I would argue as follows:

Take your loop $\gamma$, then construct a totally real submanifold $L$ that contains $\gamma$.  (If $\gamma$ is embedded the construction of $L$ does not pose any difficulty -- we are not requiring that $L$ is closed or anything similar).

Assume now that $\gamma$ bounds a *smooth* map $f\colon \Sigma \to (M,J)$ for some Riemannian surface $\Sigma$, then you can use the Riemann-Roch formula to compute the "expected" dimension of the space of holomorphic curves that are homotopic to $f$.  Assuming that $\gamma$ is injective, and that $J$ is chosen "generically" (which of course is a bit of a mysterious property), you can assume that the expected dimension corresponds to the genuine dimension of the space of holomorphic curves.

If the expected dimension is negative, then there will not be any holomorphic curve bounded by $\gamma$ in this homotopy class.
(Note that morally the higher the genus of the surface the more negative the dimension!! For example, if $S_1,S_2$ are Riemann surfaces and the genus of $S_2$ is $\ge 2$, then the expected dimension of a holormorphic curve 
in $S_1\times S_2$ that is homotopic to $\{p\}\times S_2$ is negative.  Obviously 
if we take the product almost complex structure of $S_1\times S_2$ then the manifold will be foliated by the holomorphic curves $\{p\}\times S_2$, which seems to be a contradiction to what I wrote, but this is because the almost complex structure $j_1\oplus j_2$ is highly non-generic ... as soon as you slightly perturb it, no holomorphic curves in that homotopy class will survive.)

This should solve in many cases your question (modulo "genericity" of the almost complex structure, which is not a very user-friendly definition, because you can never check if "your" $J$ is actually "generic".  But this is the way that symplectic topology goes...).

Note that up to here, we have not used that $M$ is symplectic, but only that it is almost complex.

Now we might think that in some cases the expected dimension might be positive, but even in that case, using Gromov compactness we will be able to prove generic non-existence if we perturb $\gamma$:  If there is a holomorphic curve $u$ in a certain homotopy class, we know that the nearby holomorphic curves will lie in a finite dimensional space.  It follows that only a finite dimensional family of loops close to $\gamma$ can be the boundary of a holomorphic curve that lies *close* to $u$ (in the space of smooth maps).

Unfortunately, we could still imagine that there is an infinite collection of holomorphic curves with boundary on $\gamma$.  To analyze that situation more in detail, one would have to use Gromov compactness to prove that the space of holomorphic curves sitting on $u$ is only finite, but I'll stop here.