I have a quick question: if $\mathbf{f}:\;\mathbb{R}^3\to\mathbb{R}^3$ is odd, in the sense that $\mathbf{f}(-\mathbf{v})=-\mathbf{f}(\mathbf{v})$ for any $\mathbf{v}\in\mathbb{R}^3$, and $S$ is a finite surface that is symmetric through the $xy,\,yz$ and $xz$ planes. Does:
$$\int_S \mathbf{f}\cdot \mathrm{d}\mathbf{S}=0?$$
No. Take for instance $f(x,y,z) = (x,y,z)$ and $S$ the unit sphere.
To get the property you want, you will need $f$ even: $f(-q) = q.$ Then $$\int_S f(q)\cdot n(q)\,dS = \int_S f(-q)\cdot n(-q)\,dS = \int_S f(q)\cdot -n(q)\,dS = -\int_S f(q)\cdot n(q)\,dS.$$