Consider the three cases K=0, K>0 and K<0 in your second equation. In each case, you can use ODE theory (and the values of E, F and limits of G) to solve for G, which in all cases is independent of theta. Plug in the first fundamental form coefficients into your first equation for the geodesic curvature.
First of all, when you ask "Is it correct to assume that [the integral is always zero]," you're asking the wrong question. You can assume something if it's a hypothesis, or if it's an axiom. But if you assume something in addition to the theorem's hypotheses, you're proving a different theorem.
I think what you meant to say is "Can we prove that the integral is always zero?" You offered a possible proof -- because the integral around a closed geodesic must be zero, if the integral is always the same it must be zero. But as @Ted pointed out, this argument would only be valid if you could show that $S$ always admits a closed geodesic. Think about the case in which $S$ is a plane -- in that case, there are no closed geodesics on $S$, so your argument would not be valid.
The way to approach this is to think about some special curves that always do exist. (When you know something to be true "for all $x$," it's often most useful to apply it to some specially chosen $x$ that are easy to analyze.)
In this case, one class of curves to which you can apply the hypothesis is small closed curves (with no angles) bounding regions homeomorphic to disks. (You can always find plenty of such curves by taking small circles in the plane, and looking at their images under parametrizations.) For these curves, you can explicitly evaluate every term in the Gauss-Bonnet formula. It turns out that Euler characteristic term is always the same for such curves (what is it?), and the angle-sum term is always zero, so as you suggested in your comment, the fact that the boundary integral is always the same constant implies that the integral of Gaussian curvature is always the same constant: let's call it $C$.
To determine what the constant is, consider a sequence of small circles containing a point $p$ whose radius and area go to zero. Since each integral is equal to $C$, the integrals approach $C$ as a limit. On the other hand, since the curvature is bounded near $p$ and the area is going to zero, the limit must be zero. Ergo we conclude $C=0$.
Now we've shown that $K$ has the property that its integral over every region bounded by a simple closed curve is zero. What function has that property?
Edit: For extra credit, determine what constant $\int_C \kappa_g\, ds$ is always equal to, and exhibit an example of a surface that satisfies the hypotheses.
Best Answer
Let $\alpha$ be a unit-speed parametrization of $C$ with $\alpha(0) = p$. The projection of $\alpha$ in $T_pS$ along $N(p)$ is given by $$\beta(s) = \alpha(s)+\langle p-\alpha(s),N(p)\rangle N(p),$$whence $\beta'(s) = \alpha'(s) - \langle \alpha'(s),N(p)\rangle N(p)$ and $\beta''(s) = \alpha''(s) - k_n(\alpha'(s)) N(p)$. For $s=0$ we obtain $$\beta'(0) = \alpha'(0) \quad\mbox{and}\quad \beta''(0) = \frac{D\alpha'}{{\rm d}s}(0).$$Computing the curvature of $\beta$ at $0$ we have $$\kappa_\beta(0) = \frac{\|\beta'(0) \times \beta''(0)\|}{\|\beta'(0)\|^3} = \left\|\frac{D\alpha'}{{\rm d}s}(0)\right\| \sin \theta_1,$$where $\theta_1 = \angle(\alpha'(0), (D\alpha'/{\rm d}s)(0))$. On the other hand, the geodesic curvature of $\alpha$ at $p$ is given by: $$k_g(p) = \left\langle \frac{D\alpha'}{{\rm d}s}(0), N(p) \times \alpha'(0)\right\rangle = \left\|\frac{D\alpha'}{{\rm d}s}(0)\right\|\| N(p)\times \alpha'(0)\|\cos \theta_2 = \left\|\frac{D\alpha'}{{\rm d}s}(0)\right\|\cos \theta_2 $$since $N(p)$ and $\alpha'(0)$ are orthogonal unit vectors, where $\theta_2 = \angle((D\alpha'/{\rm d}s)(0),N(p)\times \alpha'(0))$. Since $\theta_1+\theta_2 = \pi/2$, the result follows.