In general, I don't think so. If my interpretation is correct (see my comment above), then if you take an Einstein manifold with $Ric = \lambda g$ you immediately see that the best you can expect is $O(r^{2})$. In fact, if you take the standard sphere $\mathbb{S}^n$, the Taylor expansion is precisely $r^2 + \ldots$ which suggests that the Laplacian is exactly $n\neq 0$. 

It is late here so I don't feel like doing the computation, but thinking about it a bit more it seems to me that (meaning, I am making an educated guess) what you wrote down should be in fact equal to the scalar curvature at $x$.