You are confusing Stokes's theorem in $\mathbb{R}^3$ with Gauss's. You do not need a closed surface in order to apply Stokes's theorem, quite on the contrary: if you had a closed surface its boundary would be empty and the integral would be zero.
(If you're not convinced, think this way: we have a closed surface and we can apply Gauss's theorem. Therefore, we obtain
$$\iint\limits_{S} (\nabla \times \vec{F} ) \cdot d \vec{S} = \iiint\limits_{V} \nabla \cdot (\nabla \times \vec{F} ) \, dV,$$
but we have the vector identity $\nabla \cdot (\nabla \times \vec{F}) = 0,$ or $\text{div} (\text{rot}(\vec{F})) = 0$, as desired.)
That said, the boundary of your surface is pretty easy: it's the circle with radius 2 in the plane $z=0$, which can be parametrized as $\vec{r}(t) = (x(t),y(t),z(t)) = (2 \cos t, 2 \sin (t), 0).$ We can now use Stokes's theorem, yielding
$$
\begin{align}
\iint\limits_{S} (\nabla \times \vec{F}) \cdot d \vec{S} & = \int\limits_{C} \vec{F}(\vec{r}(t)) \cdot d \vec{r}(t) \\
& = \int_0^{2 \pi} (- 2 \sin (t)) \cdot (-2 \sin (t)) \, dt \\
& = \int_0^{2 \pi} 4 \sin^2 (t) \, dt \\
& = 4 \cdot \int_0^{2 \pi} \sin^2 (t) \, dt = 4 \pi.
\end{align}
$$
While Daniel Fischer's answer probably is more direct, I find this way of showing that the circulation over a closed surface is zero (using Stokes theorem) more intuitive:
Here I split the closed surface $S$ into two surfaces $S_1$ and $S_2$ with a shared boundary, the curve $C_1$. If we apply Stokes theorem to the two surfaces separately we get:
$$\int{\int_{S_1} (\nabla \times \bar F) d \bar S} = \int_{C_1} {\bar F \cdot d \bar r}$$
and
$$\int{\int_{S_2} (\nabla \times \bar F) d \bar S} = - \int_{C_1} {\bar F \cdot d \bar r}$$
notice the negative sign before the right hand integral in the second equation. This is because the curve $C_1$ runs in the opposite direction to the normals of $S_2$ compared to $S_1$.
Combing the two surface integrals to get the integral over the entire surface,
$S = S_1 \cup S_2$:
$$
\begin{eqnarray}
\int{\int_{S} (\nabla \times \bar F) d \bar S} &=& \int{\int_{S_1} (\nabla \times \bar F) d \bar S} + \int{\int_{S_2} (\nabla \times \bar F) d \bar S} \\
\int{\int_{S} (\nabla \times \bar F) d \bar S} &=& \int_{C_1} {\bar F \cdot d \bar r} - \int_{C_1} {\bar F \cdot d \bar r} \\
&=& 0
\end{eqnarray}
$$
Best Answer
Yes you can use Stokes theorem but as well you can use Gauss(divergence) theorem
$$\iint_S (\nabla \times \vec{F}) \cdot \hat{n} dS =\iiint_{\text{Interior}(S)} \nabla \cdot ( \nabla \times \vec{F}) dV$$
But divergence of curl is identically zero ie
$$ \nabla \cdot ( \nabla \times \vec{F}) = 0$$