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 R http://latex.mathoverflow.net/png?A be a regular local ring and M http://latex.mathoverflow.net/png?M a finitely generated reflexive R http://latex.mathoverflow.net/png?A-module. Then if
Ext\sb R^1(Hom(M,M),R) = 0 http://latex.mathoverflow.net/png?Ext%5FA%5E1%28Hom%28M%2CM%29%2CA%29%20%3D%200
M is free over R http://latex.mathoverflow.net/png?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 http://latex.mathoverflow.net/png?2 (more vanishing though). This I guess is not terribly surprising since the obstruction in your construction lives in codimension \geq 2 http://latex.mathoverflow.net/png?%5Cgeq%202. Indeed, with the hypothesis that A http://latex.mathoverflow.net/png?A be a UFD one has
codim; Supp\sb A(Ext\sb A^1(U,A)) http://latex.mathoverflow.net/png?codim%5C%3B%20Supp%5FA%28Ext%5FA%5E1%28U%2CA%29%29 \geq 2 http://latex.mathoverflow.net/png?%5Cgeq%202
I've been trying to do a bit better (or find a counterexample) by assuming that A http://latex.mathoverflow.net/png?A is at least Gorenstein of finite Krull dimension and using the fact that one can write down the injective resolution for A http://latex.mathoverflow.net/png?A but I haven't had any luck yet.