You are mixing up two different things; the surface integral is not a generalization of the line integral.* This is easiest to see in two dimensions, where everything is an integral along curves and yet you will still find a difference.
For simplicity, let's consider a constant wind field blowing to the right, $\mathbf f(x,y)=(1,0)$. Also consider two curves, $A$ a horizontal line segment from $(0,0)$ to $(1,0)$, and $B$ a vertical line segment $B$ from $(0,0)$ to $(0,1)$. In 2D, given a vector field and a curve there are two different kinds of integral you can consider.
Interpret the curve as a wire on which a bead is threaded. If you move the bead from one end to the other, how much does the wind help or hinder the motion of the bead? This is the usual line integral $\int \mathbf f\cdot\mathrm d\mathbf r$. It is large for curve $A$ and zero for curve $B$.
Interpret the curve as a butterfly net being held stationary while the wind blows through it. How much air passes through it per unit time? This is the flux integral $\int \mathbf f\cdot\mathbf n\,\mathrm d\ell$, where $\mathbf n$ is the unit vector perpendicular to the curve tangent. It is zero for curve $A$ and large for curve $B$.
These two notions generalize to higher dimensions in different ways. The line integral remains an integral over a $1$-dimensional object, i.e. a curve. The flux integral becomes an integral over a over an $(n-1)$-dimensional object, i.e. a surface.
*In an abstract sense one could argue that they are both specializations of the same thing, but that will take us too far into the theory of differential forms.
Your first calculation is off. We do indeed have $\nabla \times \textbf{F} = \textbf{i}+\textbf{j}+\textbf{k}$ but when we parametrize this "quarter-sphere" as
$$\textbf{X}(\phi,\theta)=(\cos \theta \sin \phi , \sin \theta \sin \phi , \cos \phi), \phi \in [0,\pi/2], \theta \in [0,\pi]$$
we have normal vector $\textbf{N}(\phi,\theta) = (\sin^2 \phi \cos \theta, \sin^2 \phi \sin \theta, \cos \phi \sin \phi)$ so evaluating the integral gives us:
$$
\begin{align}
\int_{0}^{\pi} \int_{0}^{\pi/2} \left[\sin^2 \phi (\cos \theta + \sin \theta) + \cos \phi \sin \phi \right]\, d\phi \, d\theta &= \frac{1}{4} \int_{0}^{\pi} \left(\pi \sin \theta + \pi \cos \theta + 2 \right)\, d\theta = \pi.
\end{align}
$$
Now, to use Stoke's theorem, we need a closed boundary so we can parametrize the boundary piecewise as $\textbf{x}_1 \cup \textbf{x}_2$ where
$$
\textbf{x}_1(\theta) = (\cos \theta, \sin \theta, 0) \quad \theta \in [0,\pi] \\
\textbf{x}_2(\theta) = (\cos \theta, 0, \sin \theta) \quad \theta \in [\pi,0].
$$
Evaluating a piecewise line integral gives
$$
\begin{align}
\int_{0}^{\pi} (0,\cos \theta,\sin \theta) \cdot (-\sin \theta, \cos \theta, 0) \, d\theta \, +
\\ \int_{\pi}^{0} (\sin \theta, \cos \theta, 0) \cdot (-\sin \theta, 0, 0) \, d\theta &= \int_{0}^{\pi} \cos^2 \theta + \sin^2 \theta \, d\theta \\
&= \pi.
\end{align}
$$
Best Answer
For the first integral you can use Stokes' Theorem directly and compute the surface integral over a surface $M$ as a line integral over the boundary $\partial M$ (properly oriented):
$$\iint_M (\nabla \times F) \cdot \hat{n} d\sigma = \int_{\partial M} F\cdot \hat{T} ds$$
For the second, you have to find a vector potential for $F$ - that is, to express $F$ as $\nabla \times G$ for some to-be-determined-by-you vector field $G$: $$\iint_M F\cdot \hat{n} d\sigma = \iint_M (\nabla \times G) \cdot \hat{n} d\sigma = \int_{\partial M} G\cdot \hat{T} ds$$
So:
"Can we use Stokes's Theorem to calculate the flux of a vector field across a surface?" Yes, if you find a vector potential for the given vector field. Since the divergence of a curl is zero, that would not be possible if the divergence of $F$ were not zero.
"Can we use a surface integral to calculate the flux of the curl across a surface in the direction of the normal vector?" Yes, but the computation would likely be simplified by using Stokes' Theorem - hence computing a line integral instead of a surface integral.