In the case of regular local rings a criterion for a reflexive module to be free is given here (academic access required). The result is as follows
Let $A$ be a regular local ring and $M$ a finitely generated reflexive $A$-module. Then if $\operatorname{Ext}_A^1(\operatorname{Hom}(M,M),A) = 0$ $M$ is free over $A$. This is not great though since we have just asked for cohomological vanishing elsewhere.
A criterion is also given in terms of unmixed ideals of height $2$ (more vanishing though). This I guess is not terribly surprising since the obstruction in your construction lives in codimension $\ge 2$. Indeed, with the hypothesis that $A$ be a UFD one has $\operatorname{codim} \operatorname{Supp}_A(\operatorname{Ext}_A^1(U,A)) \ge 2$
I've been trying to do a bit better (or find a counterexample) by assuming that $A$ is at least Gorenstein of finite Krull dimension and using the fact that one can write down the injective resolution for $A$ but I haven't had any luck yet.
