Let me try to prove it.

Let $M$ be an $m$-dimensional Riemannian manifold.
Set 
$$f_p(z)=\frac{|p-z|_M^2}2,$$ 
where $|p-z|_M^2$ denotes the distance from $p$ to $z$ in $M$.

Note that $M$ has nonnegative Ricci curvature if and only if 
$$(\Delta f_p)(x)\le m$$
for any $p$ and $x\in M$.
The later means that for the integral
$$\int\limits_{B_r(x)}f_p(z)\cdot d_z\mathrm{vol}$$
the comparison inequality holds, i.e., if $\tilde p,\tilde x\in\mathbb{E}^m$ and $f_{\tilde p}(\tilde x)=f_{p}(x)$ then
$$\int\limits_{B_r(x)\subset M}f_p(z)\cdot d_z\mathrm{vol}
\le 
\int\limits_{B_r(\tilde x)\subset \mathbb{E}^m}f_{\tilde p}(\tilde z)\cdot d_{\tilde z}\mathrm{vol}.$$

The last inequality survives in measured Gromov--Hausdorff limit.