This problem is setup for spherical coordinates, so I would recommend you don't use the surface area formula you've written in terms of Cartesian coordinates.
Since $r=2$ is fixed, the infinitesimal area element in spherical coordinates is $dA=4\sin\theta d\theta d\phi$. The spherical cap is bounded by $0<\theta<\pi/4$ and $0<\phi<2\pi$. So try to evaluate the integral:
$$\int_0^{2\pi}\int_{0}^{\frac{\pi}{4}}4\sin\theta d\theta d\phi$$
I will solve this problem again using the suggestion by Ted Shifrin in the comments to this question above. Doing the integral in polar coordinates, the solution becomes much simpler.
I will take the polar plane to be the $yz$ plane.
In $yz$ coordinates, the desired surface is the function $x=f(y,z)=\sqrt{4-y^2-z^2}+2$ (which corresponds to the right hemisphere of the sphere represented in the picture in the question).
As required by the formula, the surface $x=f(y,z)=\sqrt{4-y^2-z^2}+2$ is above a region $R$ which lies on the $yz$ plane (polar plane). In this case, the region of integration $R$ is the projection of the sphere on the polar plane; in other words, $R$ is a circle of radius 2 centered at $(x,y,z)=(2,0,0)$.
Thus, the limits of integration in polar coordinates will be:
$$\int_0^2\int_0^{2\pi} \cdots r d\theta dr$$
The partial derivatives of $f(x,y,z)$ are $f_y = -\dfrac{y}{\sqrt{4-y^2-z^2}}$ and $f_z = -\dfrac{z}{\sqrt{4-y^2-z^2}}$.
The square root inside the double integral is:
$$\sqrt{f_y^2(y,z)+f_z^2(y,z)+1} = \sqrt{ \dfrac{y^2}{4-y^2-z^2} + \dfrac{z^2}{4-y^2-z^2} + 1 }$$
$$= \sqrt{ \dfrac{4}{4-y^2-z^2} }$$
Converting the above expression to polar coordinates (substituting $y^2+z^2=r^2$):
$$\sqrt{ \dfrac{4}{4-r^2} }$$
Thus, the double integral in polar coordinates is:
$$\int_0^2\int_0^{2\pi} \sqrt{ \dfrac{4}{4-r^2} } r d\theta dr$$
Solving the inner integral, the above expression reduces to $\int_0^2 4 \pi r \sqrt{ \dfrac{1}{4-r^2} } dr$, which can be solved by substitution (for example, by setting $u=4-r^2$ and $du=-2rdr$) and gives $8\pi$ as its result.
Best Answer
Let us make the computation, without seeing the result as in Sabyasachi's answer, and for a general $r$.
The elementary area is, with Mathematica notations in spheric coordinates: $$ r^2\sin\phi d\phi d\theta $$
Here $r$ is a constant and $\theta\in[0,2\pi]$ and $\theta\in[0,\frac\pi 2]$:
$$ A = r^2\int_0^{\frac \pi 2}\sin \phi d\phi\int_0^{2\pi} d\theta = 2\pi [-\cos\phi]_0^{\frac \pi 2}r^2 = 2\pi r^2 $$