Answer
$\iint_S F \cdot dS=\iint_D [P \dfrac{\partial h}{\partial x}-Q+R\dfrac{\partial h}{\partial z}]dA $
where, $D$ is a projection of $S$ onto the xz-plane.
Work Step by Step
Since, $\iint_S F \cdot dS=\iint_S F \cdot n dS$
where, $n$ denotes the unit vector.
and $dS=n dA=\pm \dfrac{(r_u \times r_v)}{|r_u \times r_v|}|r_u \times r_v| dA=\pm (r_u \times r_v) dA$
The flux through a surface can be defined only when the surface is orientable.
Here, we have $r_x=i+(\dfrac{\partial h}{\partial x})j;r_z=(\dfrac{\partial h}{\partial z})j+k$
Now, we get $r_x \times r_z=(\dfrac{\partial h}{\partial x})i-j+(\dfrac{\partial h}{\partial z})k$
Thus, we get $dS=(\dfrac{\partial h}{\partial x} i-j+\dfrac{\partial h}{\partial z}k) dA$
When $F=Pi+Qj+Rk$
Hence, our result is: $\iint_S F \cdot dS=\iint_D [P (\dfrac{\partial h}{\partial x}) -Q+R(\dfrac{\partial h}{\partial z})]dA $
where, $D$ is the projection of $S$ onto the xz-plane.