In Griffiths introduction to Electrodynamics, the classic image problem is presented: There is a charge q, above a grounded conducting plane.
The boundary conditions are therefore: 1.V=0, at plane 2.v=0 at infinity
My question is, since potentials are harmonic functions and the potential is zero both inside and on boundary, shouldn't the potential therefore be zero everywhere in R^3 ?
No, because the potential in this case in not an harmonic function.
Your statement was correct if the differential equation you have had to solve was the Laplace's equation $$\nabla^2V=0$$ which accounts for space without charges. In that case the potential was indeed an harmonic function, and in order to satisfy the maximum principle - it must have vanish identically.
However, in your particular case there are charges in the volume of interest, and thus one needs to fulfill the Poisson's equation $$\nabla^{2}V=-\frac{Q}{\varepsilon_{0}}\delta\left(x\right)\delta\left(y\right)\delta\left(z-d\right)$$ in the upper half part of $\mathbb{R}^{3}$. Here I assumed the charge $Q$ is placed at $\vec{r}=\left(0,0,d\right)$. Therefore, the function is not harmonic and the maximum principle is not applicable.
