First, the relation $R(x,y)$ as you defined it is in general not type-definable (although it is definable in any $\aleph_0$-categorical theory, since it is invariant). Indeed, take any stable structure in which $X = \mathrm{acl}(\emptyset)$ is infinite, and take any sequence $(a_n) \subseteq X$ without repetitions. Then we have $R(a_n,a_n)$ for all $n$, but in an ultra-power, if $a = [a_n]$ (the class of the sequence modulo the ultra-filter) then $a \notin \mathrm{acl}(\emptyset)$, so $\neg R(a,a)$. This happens, for example, in the theory of algebraically closed fields, which is as stable as you can get.

Second, for any fixed $b$, the relation $R(x,b)$ is type-definable: $$ R(x,b) = \{ \neg \varphi(x,b) \colon \text{the formula } \varphi(x,b) \text{ forks over } \emptyset \}. $$ I stated this over the empty set, as in the question, but the same is true over an arbitrary parameter set.

The problem is that as $b$ varies, the property "$\phi(x,b)$ forks over $\emptyset$" varies in a non type definable fashion, so you really must know $b$

First, the relation $R(x,y)$ as you defined it is in general not type-definable (although it is definable in any $\aleph_0$-categorical theory, since it is invariant). Indeed, take any stable structure in which $X = \mathrm{acl}(\emptyset)$ is infinite, and take any sequence $(a_n) \subseteq X$ without repetitions. Then we have $R(a_n,a_n)$ for all $n$, but in an ultra-power, if $a = [a_n]$ (the class of the sequence modulo the ultra-filter) then $a \notin \mathrm{acl}(\emptyset)$, so $\neg R(a,a)$. This happens, for example, in the theory of algebraically closed fields, which is as stable as you can get.

Second, for any fixed $b$, the relation $R(x,b)$ is type-definable: $$ R(x,b) = \{ \neg \varphi(x,b) \colon \text{the formula } \varphi(x,b) \text{ forks over } \emptyset \}. $$ I stated this over the empty set, as in the question, but the same is true over an arbitrary parameter set.

The problem is that as $b$ varies, the property "$\varphi(x,b)$ forks over $\emptyset$" varies in a non type definable fashion, so you really must know $b$