Let $R$ be a commutative ring with unity. Let $T: R$-Mod $\to R$-Mod be an additive, covariant, left exact functor which commutes with direct limits indexed by directed sets. Let $R^i T$ be the right derived functors of $T$.
Is it true that for any directed system of modules $\{M_{\alpha}\}$ indexed by a directed set, we have $\varinjlim (R^i T) (M_\alpha)\cong (R^iT)(\varinjlim M_\alpha)$ ?
No. Indeed, this need not even be true for $T$ itself (which is $R^0T$). For instance, if $I$ is an infinite set (and $R$ is nonzero), the functor $T(M)=M^I$ is exact but does not preserve directed colimits, since if $M$ is the direct limit of $(M_\alpha)$ then there may be elements of $M^I$ whose coordinates do not all come from any single $M_\alpha$.
Even if $T$ preserves directed colimits, its derived functors may not. For instance, let $A=k[S]$ be a polynomial ring over a field $k$ with an infinite set $S$ of variables, let $I=(S)$ be the ideal generated by all the variables, and let $R=A/I^2$. Let $N=R/(s)$ for some variable $s\in S$ and consider the functor $T=\operatorname{Hom}(N,-)$. Since $N$ is finitely presented, $T$ preserves directed colimits. To compute the derived functors $R^iT=\operatorname{Ext}^i(N,-)$ we take a minimal free resolution of $N$ which has the form $$\to R^{\oplus S}\to R\stackrel{s}\to R\to N\to 0$$ (where "minimal" means every map in the resolution is $0$ mod $I$). If $M$ is any $R$-module which is annihilated by $I$, we then see that $$\operatorname{Ext}^2(N,M)\cong \operatorname{Hom}(R^{\oplus S},M)\cong M^S.$$ Since $S$ is infinite, $M\mapsto M^S$ does not preserve directed colimits, so $R^2T=\operatorname{Ext}^2(N,-)$ does not preserve directed colimits.
If you assume that $T$ preserves directed colimits and $R$ is Noetherian, then it is true. As a sketch of a proof, if $M$ is a colimit of a directed system $(M_\alpha)$, then we can construct an injective resolution of $M$ as a directed colimit of injective resolutions of the $M_\alpha$, using the fact that directed colimits of injective modules are injective since $R$ is Noetherian. (This step is nontrivial, since we can't actually cobble injective resolutions of all the $M_\alpha$ into a diagram that commutes on the nose in any obvious way. One way to handle this is to reduce to the case that the system $(M_\alpha)$ is indexed by an ordinal and is cocontinuous, so you can build a commutative diagram of injective resolutions by transfinite induciton.) We then see that computing $R^iT(M)$ using this injective resolution is the same as computing $R^iT(M_\alpha)$ using their injective resolutions and then taking the colimit.