Here is an extended version of a comment above.

First, one can consider $R\Gamma\Omega^\bullet_X$ as a complex of vector spaces. Any complex of vector spaces is quasi-isomorphic to its cohomology. Taking this complex to be $End(R\Gamma\Omega^\bullet_X)$ we get
$$End(R\Gamma\Omega^\bullet_X)\sim H^\bullet (End(R\Gamma\Omega^\bullet_X))\cong End (H^\bullet(X)).$$

Here $\sim$ stands for quasi-isomorphic and we use the fact that if $K^\bullet$ and $L^\bullet$ are two complexes of vector spaces, then (as graded vector spaces) $$H^\bullet (Hom(K^\bullet, L^\bullet))\cong Hom (H^\bullet(K^\bullet), H^\bullet(K^\bullet)).$$

There is one slightly tricky point here: both $End(R\Gamma\Omega^\bullet_X)$ and $ End (H^\bullet(X))$ are dg-algebras (the latter with zero differential) and one can ask if they are quasi-isomorphic as such. The answer is yes; this can be shown as follows: if $K^\bullet$ and $L^\bullet$ are homotopy equivalent complexes (in fact, in any abelian category), there is an $A_\infty$ quasi-isomorphism between $End(K^\bullet)$ and $End(L^\bullet)$ and hence $End(K^\bullet)$ and $End(L^\bullet)$ are quasi-isomorphic as dg-algebras, meaning they can be connected by a chain of quasi-isomorphisms. (This is probably an overkill, but I can't think of a more straightforward argument at the moment.)

Second, e.g. by taking the Dolbeault resolution one can think of $R\Gamma\Omega^\bullet_X$ as a commutative dg-algebra (cdga) and so one can consider $End_{R\Gamma\Omega^\bullet_X}(R\Gamma\Omega^\bullet_X)$, the endomorphisms of $R\Gamma\Omega^\bullet_X$ as a module over itself. The result will be again $R\Gamma\Omega^\bullet_X$: we use the fact that the natural identification of algebras $Hom_A(A,A)\cong A$ is compatible with the differentials if $A$ is a dg-algebra. Now, again from the Dolbeault resolution, $R\Gamma\Omega^\bullet_X$ is just the algebra of smooth complex-valued forms on $X$, which, as shown by Deligne, Griffiths, Morgan, SUllivan, Real homotopy theory... is quasi-isomorphic to $H^\bullet (X)$ with zero differential.


So we get $$End_{R\Gamma\Omega^\bullet_X}(R\Gamma\Omega^\bullet_X)\cong  R\Gamma\Omega^\bullet_X\sim H^\bullet (X)\cong End_{H^\bullet(X)}(H^\bullet (X)).$$