What you need is the Stoke's formula, which asserts that $\int_{\partial S}w=\int_{S}dw$ for a bounded open manifold $S$. Therefore we have $\int_{\partial S}wdz=\int_{S} dw\wedge dz=\int_{S}\frac{\partial w}{\partial \overline{z}}dz\wedge d\overline{z}$ using the fact that $dw=\frac{\partial w}{\partial \overline{z}}d\overline{z}+\frac{\partial{w}}{\partial z}dz$
Thus for your example we have $\int_{\partial S}\overline{z}^{3}=3\int_{S}\overline{z}^{2}dz\wedge d\overline{z}$
This integral seems can be perfectly non-zero in appropriate regions. I am not sure how the integral works in practice (say, giving a retangle or a circle) without breaking into real and imaginary components. We at least have $dz\wedge d\overline{z}=(dx+idy)\wedge (dx-idy)=-idx\wedge dy+idy\wedge dx=-2idx\wedge dy$ While $\overline{z}^{2}=x^{2}-y^{2}-2ixy$
Thus separating the real and imaginary components we should have $-12\int_{S}xydxdy-6i\int_{S}(x^{2}-y^{2})dxdy$
The real and complex part can easily be nonzero I guess.