Gauss theorem and inverse square law I know that the gauss law states that the Flux of the electric field through a closed surface is Q/ε , but does the gauss theorem works also for non inverse square law Fields? 
 A: I'd like to draw a distinction: 
Gauss's Theorem (Also called Divergence Theorem): 
$$\iint_{\partial V} \mathbf{E} \cdot d\mathbf{A} = \iiint_V \nabla \cdot \mathbf{E}\ dV $$
This is a purely mathematical statement and holds for all differentiable vector fields $\mathbf{E}$. 
Gauss's Law: $$\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0}$$
Plugging this into Gauss's Theorem we have that 
$$\iint_{\partial V} \mathbf{E} \cdot d\mathbf{A}= \frac{1}{\epsilon_0} \iiint_V \rho(\mathbf{r}) \ dV \equiv \frac{Q_{enc}}{\epsilon_0}. $$
So to answer your question, Gauss's Theorem is always true. It must be. However, Gauss's Law didn't have to be true; it just so happens to be a law of physics in the universe we find ourselves in. 

Having said that, Gauss's law will be true for any vector field $\mathbf{F}$ that satisfies the differential equation.
$$\nabla \cdot \mathbf{F} \propto \Lambda(\mathbf{r}), $$
where $\Lambda$ is just some scalar field that is well defined over the volume $V$. “Gauss's law” (i.e. in integral form; I put it in quotes because we are just plugging it into Gauss’s theorem, but this is how many use the term) for such a field would look like 
$$\iint_{\partial V} \mathbf{F} \cdot d\mathbf{A}= \alpha \iiint_V \Lambda(\mathbf{r}) \ dV \equiv \alpha \tilde{Q}_{enc}, $$
for some constant $\alpha$ and $\tilde{Q}_{enc}$ is just what we define to be how much "charge" is enclosed by our surface.
Conclusion: The fact that the $\mathbf{E}$ field falls like $\frac{1}{r^2}$ just makes the integral on the left hand side do-able at a fixed $r$ (actually the fact that it only depended on $r=|\mathbf{r}|$ is what made it do-able); it does not mean that anything that the corresponding field $\mathbf{F}$ falls like $\frac{1}{r^2}$. When you think about all the times you actually used Gauss's law to calculate the $\mathbf{E}$ field recall that you actually had to assume that it fell like $\frac{1}{r^2}$ in order to do the integral simply. 
