I'll add another example to the mix even in the ring setting (as opposed to the sheaf setting). Fix a DVR $(R, \langle x \rangle)$. Then the fraction filed $K(R) = \bigcup_n (x^{-n}R)$, is an ascending union of free (reflexive) modules. But clearly $K(R)$ is not itself reflexive.

On the other hand, under mild conditions on the scheme $X$, for example if it is normal, then for a coherent module being reflexive is equivalent to being S2 (Serre's second condition). The S2 condition behaves fairly well for quasi-coherent modules and might be useful for you. For example, see the paper(s) of Hartshorne: "Generalized divisors ..."

I'll add another example to the mix even in the ring setting (as opposed to the sheaf setting). Fix a DVR $(R, \langle x \rangle)$. Then the fraction field $K(R) = \bigcup_n (x^{-n}R)$, is an ascending union of free (reflexive) modules. But clearly $K(R)$ is not itself reflexive.

On the other hand, under mild conditions on the scheme $X$, for example if it is normal, then for a coherent module being reflexive is equivalent to being S2 (Serre's second condition). The S2 condition behaves fairly well for quasi-coherent modules and might be useful for you. For example, see the paper(s) of Hartshorne: "Generalized divisors ..."

I'll add another example to the mix even in the ring setting (as opposed to the sheaf setting). Fix a DVR $(R, \langle x \rangle)$. Then the fraction filed $K(R) = \bigcup_n (x^{-n}R)$, is an ascending union of free (reflexive) modules. But clearly $K(R)$ is not itself reflexive.

On the other hand, under mild conditions on the scheme $X$, for example if it is normal, then being a reflexive coherent module is equivalent to being S2 (Serre's second condition). The S2 condition behaves fairly well for quasi-coherent modules and might be useful for you. For example, see the paper(s) of Hartshorne: "Generalized divisors ..."

I'll add another example to the mix even in the ring setting (as opposed to the sheaf setting). Fix a DVR $(R, \langle x \rangle)$. Then the fraction filed $K(R) = \bigcup_n (x^{-n}R)$, is an ascending union of free (reflexive) modules. But clearly $K(R)$ is not itself reflexive.

On the other hand, under mild conditions on the scheme $X$, for example if it is normal, then for a   coherent module being reflexive is equivalent to being S2 (Serre's second condition). The S2 condition behaves fairly well for quasi-coherent modules and might be useful for you. For example, see the paper(s) of Hartshorne: "Generalized divisors ..."