This is not true in general. For example, if $R$ is a free associative algebra of rank $>1$, then $M_n(R)$ does not have nontrivial identities for any $n$.
Moreover, I believe there are examples of algebras $R$ such that much stronger condition holds: $M_n(R)$ is isomorphic to $M_m(R)$ for some $m \ne n$. This reminds cancellation problems in the commutative setting, though I cannot provide examples of such algebras.
This is true if, for example, if $R$ is finite-dimensional and prime (even not necesary associative). Then we can note that $M_n(R)$ is prime too, pass to the algebraic closure of the base field, and invoke theorem of Razmyslov that finite-dimensional prime algebras over algebraically closed fields are determined by their identities (Yu.P. Razmyslov, Identities of Algebras and Their Representations, AMS, 1994 (translation from Russian), around p. 30).