I heard about this conjecture from Sara Billey at FPSAC, and I think I've got an argument. Let $F : \mathbb{C}[S_n] \to \mathbb{Z}[x_1, \ldots, x_N]^{S_N}$ be the linear map sending $w \mapsto p_{\rho(w)}(x_1, \ldots, x_N)$, and $V$ a complex vector space with $\dim V = N \geq n$.
Lemma: If $\alpha \in \mathbb{C}[S_n]$ acts on $V^{\otimes n}$ (on the right) with nonnegative eigenvalues, then $F(\alpha)$ is Schur-positive.
Proof: By a density argument we can assume $\alpha$ acts diagonalizably: say $V^{\otimes n} = \bigoplus_{\omega} U_{\omega}$ where $U_{\omega}$ is the $\omega$-eigenspace of $\alpha$. Each $U_{\omega}$ is a left $\operatorname{GL}(V)$-module since the left $\operatorname{GL}(V)$-action commutes with $\alpha$. Let $X \in \operatorname{GL}(V)$ have eigenvalues $x_1, \ldots, x_n$. The trace of $X \times \alpha$ on $V^{\otimes n}$ (i.e. of the map $z \mapsto Xz\alpha$) is, on the one hand, $\sum_{\omega} \omega \operatorname{tr}(X|_{V_\omega})$. Since $V_\omega$ is a $\operatorname{GL}(V)$-module, $\operatorname{tr}(X|_{V_\omega})$ is a Schur-positive polynomial in $x_1, \ldots, x_N$. On the other hand, $\operatorname{tr}(X \times \alpha) = F(\alpha)$.
Since $\mathbb{C}[S_n]$ acts faithfully on $V^{\otimes d}$, the eigenvalues of $\alpha$ acting on $\mathbb{C}[S_n]$ or on $V^{\otimes d}$ are the same, ignoring multiplicity (maybe the lemma can be modified to work directly on $\mathbb{C}[S_n]$?). Up to constant factors, $\sum_{u \in R_D} \chi(u)u \sum_{v \in C_D} \psi(v)v$ is the product of two idempotents, which are both Hermitian with respect to the inner product on $\mathbb{C}[S_n]$ for which permutations form an orthonormal basis. The product of two positive semidefinite matrices has nonnegative eigenvalues, so $F(\sum_{u \in R_D} \chi(u)u \sum_{v \in C_D} \psi(v)v)$ is Schur-positive by the lemma.
