This is not a complete answer, but it is a line of attack.

Theorem: The property $T_\mathrm{rc}$ implies $T_2$ if, and only if, $T_\mathrm{rc}$ is preserved by finite products.

Proof. ($\Leftarrow$) Suppose $X$ is $T_\mathrm{rc}$. By assumption $X \times X$ is also $T_\mathrm{rc}$, therefore its diagonal $\Delta_X = \{(x,y) \in X \times X \mid x = y\}$ is closed, as it is a retract of $X \times X$ by the map $(x,y) \mapsto (x,x)$. But a space is $T_2$ iff its diagonal is closed, so $X$ is $T_2$.

($\Rightarrow$) Suppose we had a space $X$ which were $T_\mathrm{rc}$ but not $T_2$. Then $X \times X$ is not $T_\mathrm{rc}$ because its diagonal is not closed. Thus $T_\mathrm{rc}$ is not preserved by finite products. QED.

We actually see that we only used a binary self-product $X \times X$. Your question is thus precisely as hard as: Is there is a $T_\mathrm{rc}$ space $X$ such that $X \times X$ is not $T_\mathrm{rc}$?

This is not a complete answer, just a possibly useful observation.

Theorem: If the property $T_\mathrm{rc}$ is preserved by finite products then any $T_\mathrm{rc}$ space is $T_2$.

Proof. Suppose $T_\mathrm{rc}$ is preserved by finite products and $X$ is $T_\mathrm{rc}$. By assumption $X \times X$ is also $T_\mathrm{rc}$, therefore its diagonal $\Delta_X = \{(x,y) \in X \times X \mid x = y\}$ is closed, as it is a retract of $X \times X$ by the map $(x,y) \mapsto (x,x)$. But a space is $T_2$ iff its diagonal is closed, so $X$ is $T_2$.