In the case of regular local rings a criterion for a reflexive module to be free is given <a href="http://www.springerlink.com/content/j2406t088q58n710/fulltext.pdf"> here </a> (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.