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.)

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.)