I'll assume the sphere is given by the equation $x^2+y^2+z^2 = R^2$ but the answer does not depend at all on $R$ - the answer is zero.
By the divergence theorem, the surface integral is the volume integral of the divergence:
\int_S A(x,y,z) ds = \int_V \nabla \cdot A dv = \int_V (y + z ) dv = \int_V y dv
+ \int_V z dv
where V is the volume of the sphere. Since ofr the function $z$ each point in the lower hemisphere cancels the corresponding point in the upper hemisphere, $\int_V z dv = 0$.
Similarly, $\int_V y dv = 0$.
So the volume integral of $\nabla \cdot A$ is zero and by the divergence theorem, the surface integral of $A$ is zero.