This is an effort to get this question off from the unanswered queue.
Also here I use a rather general approach than the classical Kelvin-Stokes theorem.
Stokes theorem reads:
$$
\int_{\partial M} \omega = \int_{M} d\omega.
$$
Hence
$$
\int_C y\,dx + z\,dy +x\,dz = \pm \int_S d(y\,dx + z\,dy +x\,dz) \\
= \pm\int_S dy\wedge dx + dz\wedge dy + dx\wedge dz.
$$
For simplicity we choose $S$ to be planar surface bounded by the sphere, not the hemisphere bounded by the plane. On the plane $z = -x-y$, hence above integral is
$$
\pm 3\int_S dx\wedge dy.\tag{$\star$}
$$
Using the natural parametrization of the plane, this integral can be interpreted as the area of projected area $S$ on to the $xy$-plane (or notice $\star dz = dx\wedge dy$). The intersection of $x+y+z = 0$ with $x^2 + y^2 + z^2 =a^2$, projected onto the $xy$-plane, is an ellipse.
The end points for this projected ellipse are: minor axis end points are projection of $(-1/\sqrt{6},-1/\sqrt{6},2/\sqrt{6} )a$, and $(1/\sqrt{6},1/\sqrt{6},-2/\sqrt{6})a$, achieved when we set $x=y$. We can compute the length of the minor axis is $b = \sqrt{3}a/3$. The major axis end points are achieved by setting $x+y=0$, $(\sqrt{2}/2, -\sqrt{2}/2, 0)a$ and $(\sqrt{2}/2, -\sqrt{2}/2, 0)a$, the length of the major axis is just $a$. Hence the project area is $\sqrt{3}\pi a^2/3$ and plugging back to $(\star)$ yields
$$
\int_C y\,dx + z\,dy +x\,dz = \pm \sqrt{3} \pi a^2.
$$
The sign depends on $C$ is chosen to be rotated counter-clockwisely or clockwisely with respect to the normal vector to the plane $x+y+z=0$.
To address your own question:
Why I got $\displaystyle \iint_{S} \operatorname{curl} \vec{F} \cdot \vec{n} ~dS = -3 \iint_{A} dA = -3 \pi a^2$ if I use $dS = \sqrt{3}dA$ so that $n\,dS = (1/\sqrt{3},1/\sqrt{3},1/\sqrt{3})\sqrt{dS} = (1,1,1)$?
This is not correct, for that if you use the area element $dS = \sqrt{3} \,dx dy = \sqrt{3} \,dA$, the new $A$ is the projected ellipse on the $xy$-plane, having area $\sqrt{3}\pi a^2/3$ (please see the argument above), not having the same area $\pi a^2$ with $S$.
Standard way using Kelvin-Stokes as Arturo Magidin pointed out in the comments: choosing $C$ rotates counter-clockwisely with respect to the unit vector $n = -(1,1,1)/\sqrt{3}$ normal to the plane $x+y+z=0$:
$$
\int_C y\,dx + z\,dy +x\,dz = \int_{S} \nabla \times (y,z,x)\cdot n\,dS
\\
= \int_{S} (-1,-1,-1)\cdot (-1,-1,-1)/\sqrt{3} \,dS = \sqrt{3}|S| = \sqrt{3}\pi a^2.
$$
You've chosen to put $u$ as the last variable to integrate. This is not how I would personally go about it, but let's continue.
Let's transform these points in terms of $u$ and $v$. Our points become
$$(u, v) = (0, 2), (0, 4), (-4, 4), (-2, 2).$$
Plot these on the $u$-$v$ plane. Note that $u$ ranges not from $-2$ to $0$, but $-4$ to $0$. The upper bound for $v$ is indeed $4$, but the lower bound is made up of two line segments, which have a cusp at $u = -2$. So, we should probably split the domain up.
For $u$ between $-4$ and $-2$, the lower bound follows the line $v = -u$. For $u$ between $-2$ and $0$, the line is $v = 0$. So, our integral is split like so:
$$\iint_A = \int_{-4}^{-2} \int_{-u}^4 + \int_{-2}^{0} \int_0^4.$$
If you instead put $v$ on the outside, you get something a lot nicer:
$$\int_2^4 \int_{-v}^0$$
Don't forget the Jacobean too!
Best Answer
By using polar coordinates $$\begin{gather} \begin{cases}x=\rho\cos{\varphi}, \\ y=\rho\sin{\varphi}, \end{cases}\end{gather}$$ $$\iint\limits_{x^2+y^2\leqslant {1}}2\ dx\,dy=2\int\limits_{0}^{1} \int\limits_{0}^{2\pi}{\rho} \ d{\varphi} \, d{\rho}=2\pi.$$