$C$-conjugation of a gluon

In some explanations about the OZI rule ( for example at page 38 here), I found that gluons have definite eigenvalue of the charge conjugation operator $$C$$. The eigenvalue is $$-1$$. How can this result be demonstrated?

I'm not sure how you "found" it. The off-diagonal gluons, like $$\bar R G$$, are not eigenstates, since they transform to their antigluon, here $$\bar G R$$.
The color-diagonal ones, like $$\bar R R - \bar G G$$, would naturally associate with the vector mesons with an analogous flavor structure to this color structure. I assume you have worked out the charge conjugation of the $$\rho^0$$, or the $$^3S_1$$ orthopositronium, namely −. That is, unphysical, strictly combinatorially, if your diagonal gluon "decayed" to a notional fermion-antifermion pair with these colors, its C would be −.
You don’t understand your ref's statement because it is wrong, or, at best, sloppy (and, in any case, subpar); only three gluons in a colorless state have negative C. The point is the color coupling $$ig_s G^\mu_{\bar RG}\overline{\psi_G}\gamma_\mu \psi_R + h.c.$$ (representing $$\lambda_+$$ in your p.5 conventions, unlike the diagonal $$\lambda_3$$ I used above!) should remind you of the EM coupling $$ie A^\mu\bar\psi\gamma_\mu \psi + h.c.$$ which goes to itself under C, so the photon is odd, since the current is odd; except the color coupling goes to $$\lambda_-$$, hence cannot be an eigenstate! So, the only gluons that are eigenstates of C are $$\lambda_3$$ and $$\lambda_8$$. Still, by taking the trace of three such color matrices, however, you may find a C-invariant combination of the fermion sextilinear, and so "pretend" each gluon G comes with an odd C eigenvalue, just like the photon. Your reference is "summarizing" all this in code, with a wink and a "you know", and you are within your rights to not be fully satisfied. There is no substitute to going back to your QFT text and checking the Dirac couplings defining C the normal way.