Double orthogonal complement of a finite module

Your question does not specify what $q$ is. But if $q$ is an odd prime -- so that you are talking about quadratic spaces over the field $\mathbb{F}_q$ of odd characteristic -- then the answer is yes.

In this case your inner product $\cdot$ is the bilinear form associated to the quadratic form $q(x_1,\ldots,x_n) = x_1^2 + \ldots + x_n^2$. This quadratic form is nondegenerate, so the result you want is Proposition 7 in these notes. (They are nothing so special: any sufficiently basic text on quadratic forms will contain this material.)

I am not really used to thinking about quadratic forms either in characteristic $2$ or over rings which are not domains, so if you're really interested in the case of $q$ not necessarily an odd prime, please say so, so that someone else can give a more complete answer. (But I will guess that the result is also true when $q$ is an odd prime power, for instance.)

The answer is yes. The easiest way for me is to appeal to character theory. If $\zeta$ is a complex $q$-th root of 1 then the map from $\mathbb{Z}_q^n$ to $\mathbb{C}$ given by $$ x \mapsto \zeta^{a^Tx},\qquad (a\in\mathbb{Z}_q^n) $$ is a character of the abelian group $W=\mathbb{Z}_q^n$. The set of characters obtained as $a$ varies over the elements of $W$ is the character group $W^*$ of $W$. If $V$ is a subgroup of $W$, then $V^\perp=V^*$ is isomorphic to the subgroup $(W/V)^*$ of $W^*$.

A convenient source for the relevant character theory is from our own KConrad: http://www.math.uconn.edu/~kconrad/blurbs/ (under characters of finite abelian groups).