As explained in [this answer](http://mathoverflow.net/a/81938/42091), the property $V=(V^\perp)^\perp$ holds for submodules $V$ of $(\mathbb Z/q\mathbb Z)^n$ because you can use Smith normal form to find a suitable generating matrix for $V^\perp$. If you just want a reference for the result, I can only suggest the following paper:

D. Wilding, M. Johnson, M. Kambites. Exact rings and semirings.
*Journal of Algebra*, 388 (2013), 324–337; [arXiv:1212.5358](http://arxiv.org/abs/1212.5358).

When writing this paper we faced the same problem as you. That $(V^\perp)^\perp=V$ follows quite easily from known results, but we could not find it written down explicitly in this formulation.