This question has been asked and answered on MathOverflow. I have replicated the accepted answer by Chris Godsil below.
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).