You are right, the curve $C$ is a great circle. This is rather obvious when you note that the center of the sphere is the origin and the plane passes through this point. For Stokes' theorem you need a surface $S$ such that $\partial S = C$. There are many choices but I would propose to the part of the plane that is "cut-out" by the sphere (which is a disk).
We then have that
$$\int_C (y dx +z dy + x dz) = -\int_S (dx\wedge dy + dy \wedge dz + dz \wedge dx) $$
by Stokes' theorem. The plane is parameterized by $u,v$ via
$$ (x,y,z) = (u,v,-u-v).$$
We obtain
$$ \int_C (y dx +z dy + x dz) = - \int_S [du\wedge dv + dv \wedge(-du-dv) + (-du -dv)\wedge du ] = - 3 \int_S du \wedge dv .$$
The remaining integral $\int_S du \wedge dv$ is the projection of the disk $S$ onto the xy-plane. The disk has radius $a$ and is slanted at an angle of $ \phi=\arccos(1/\sqrt{3})$ with respect to the xy-plane. Elementary geometry yields $\int_S du \wedge dv = \pi a^2 /\sqrt{3}$ and thus
$$ \int_C (y dx +z dy + x dz) = - \sqrt{3} \pi a^2$$
There are at least to ways you can parametrize curve $C$.
Method 1. Since the curve is at the intersection of the sphere and the cylinder, it belongs to the cylinder, which has radius $b$ and is shifted $b$ units along the $x$ axis, so you can write:
$$
x=b\cos\theta+b, \quad y =b\sin\theta, \quad 0\le \theta \le 2\pi,
$$
but since it also belongs to the sphere, you can write
$$
z=\sqrt{a^2-y^2-(x-a)^2}=\sqrt{a^2-(b\sin\theta)^2-(b\cos\theta+b-a)^2},
$$
which gives you a first parametrization.
Method 2. The cylinder has polar equation $r(\theta)=2b\cos\theta$, so
$$
x=r(\theta)\cos\theta=2b\cos^2\theta, \quad y =2b\cos\theta\sin\theta, \quad -\pi/2\le \theta \le \pi/2,
$$
and again
$$
z=\sqrt{a^2-y^2-(x-a)^2}=\sqrt{a^2-(2b\cos\theta\sin\theta)^2-(2b\cos^2\theta-a)^2}.
$$
Note the difference of variations of $\theta$ in the two parametrizations.
This being said, Stokes theorem is probably more appropriate for this question, so you are going to need to parametrize the surface. You could do it in spherical coordinates, but I think cartesian coordinates are the way to go here. Take $x$ and $y$ as parameters and you have:
$$
x=x, \quad y=y, \quad z=\sqrt{a^2-y^2-(x-a)^2},
$$
where $x$ and $y$ belong to the projection of the cylinder in the $xy$ plane, that is,
$$
(x,y)\in D:=\{(r,\theta)\;|\; -\pi/2\le \theta \le \pi/2,0\le r\le 2b\cos\theta \}
$$
It follows by the Stokes theorem that your integral equals
$$
\iint_D \nabla\times\vec{F}(x,y)\cdot \vec{r_x}\times \vec{r_y}\; dA,
$$
where $\nabla\times\vec{F}(x,y)$ is the curl of your field expressed as a field depending only on $x$ and $y$. I am sure you can takes computations from there.
Best Answer
You can check why the range for $t$ I provided in my comment will work by checking values of $t$ within that interval - what do you get when $t=0$? $t=\dfrac{\pi}{2}$? and so on.
The integral becomes $$\begin{align*}I&=\oint_C(y-z)\,\mathrm{d}x+(z-x)\,\mathrm{d}y+(x-y)\,\mathrm{d}z\\[1ex] &=\int_0^{2\pi}\bigg((a\sin t-b+b\cos t)(-a\sin t)+(b-b\cos t-a\cos t)(a\cos t)+(a\cos t-a\sin t)(b\sin t)\bigg)\,\mathrm{d}t\\[1ex] &=\int_0^{2\pi}\bigg(-a^2-ab+ab\cos t+ab\sin t\bigg)\,\mathrm{d}t\end{align*}$$