Provided $u$ is $C^2$ (so that the Laplacian makes sense), yes. Essentially, if $$u(x) = \frac{1}{|\partial B(x,r)|}\int_{\partial B(x,r)} u(y)\,{\rm d}S(y),$$make a change of variables $y = x+rz$ with $z \in \partial B(0,1)$, differentiate under the integral sign and change back to the original variable to obtain that $$\triangle u(x) = \frac{1}{|\partial B(x,r)|}\int_{\partial B(x,r)} \triangle u(y)\,{\rm d}S(y).$$Since $\partial B(0,1)$ has no boundary, Stokes' Theorem says that the above integral vanishes and so $\triangle u = 0$.
For the second query, yes, it would suffice in view of the above, but it is much simpler to just verify that $\triangle u =0$ directly by using the polar coordinate expression $$\triangle u = \frac{\partial^2u}{\partial r^2} + \frac{1}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2}\frac{\partial^2u}{\partial \theta^2} = 0 + \frac{\sin\theta}{r} - \frac{r\sin\theta}{r^2} = 0.$$