Why must a hadronic decay of the $J/\psi$ meson include (at least) three gluons? Why is the decay mediated by a single gluon allowed for the $\rho^0$ meson?
Gluons carry color. So a colorless meson cannot annihilate its valence quarks to decay to one colored gluon and nothing else. You are flat wrong that the ρ decays thus through one gluon.
Two colored gluons, may combine into a colorless object, think $\bar R G$ and $\bar G R$, and the singlet has C=+. But the ψ has C=- , like the neutral ρ. Neither can decay through 2 gluons.
The ψ can, and does decay through ggγ : check your PDG.
Finally the (also charmonium, but scalar) χ has C=+ , and can and does decay through gg.
It's hard to see how you'd learn about the three gluon decay without a justification of the reason.
Edit in response to comment: The possibility arises that you are not focussed on Zweig's rule violation, but are interested in a two-meson decay mode involving the original valence quarks. This mode is available to the ρ, above threshold for a 2 π decay, but not to the J/Ψ(3097), below threshold for a 2 D(1865) decay; so that channel is closed, and you must decay via annihilation of your valence quarks.
By sharp contrast, the Ψ(3770) is above threshold, and so does very much decay to 2 D s, so, then, analogously to the decay of the ρ , somewhat misleadingly dubbed in the question as a "one gluon decay". The colored quark pair created by that gluon has to combine with the (color altered) valence quarks and hadronize through a resolutely nonperturbative process involving quintillions of soft gluons.