I think the following argument settles the conjecture that $\inf |\rho_{\zeta}-\rho_{\zeta'}|=0$, assuming RH. Basically, it follows because the zeros can be arbitrarily close.
Littlewood proved that $$\inf \gamma_n-\gamma_{n-1}=0.$$ On RH this implies that $$\inf \rho_n-\rho_{n-1}=0.$$ So, for every $\epsilon>0$ one can find a pair of zeros between which $$\max\{|\zeta(1/2+it)|:\gamma_{n(\epsilon)-1}<t<\gamma_{n(\epsilon)}\}=\epsilon.$$
By the theorem of Macdonald (cited in my question above) and the inverse mapping theorem, the derivative must vanish somewhere on the boundary of the closed curve enclosing $\rho_{n(\epsilon)}$ defined by the condition that $|\zeta(s)|=\epsilon$, and not inside it. By the maximum principle, as $\epsilon\rightarrow 0$, such a curve must also become small and, since the derivative vanishes somewhere on it, the result follows.