Let $X$ be a $\mathbb{R}$-defined smooth proper scheme, and let $H^i_{\text{dR}}(X)$, denote its algebraic de Rham cohomology. The Hodge filtration gives an $\mathbb{R}$-defined pure Hodge structure on $H^i_{\text{dR}}(X)$, which yields in practice a (Hodge) filtration on this $\mathbb{R}$-vector space.

It is said that, as a part of the definition of a Hodge structure, this filtration must satisfy a certain property, namely that for every pair of non-negative integers, $p+q = i + 1$, $\mathcal{F}^pH^i_{\text{dR}}(X)\cap \overline{\mathcal{F}^qH^i_{\text{dR}}(X)} = 0$.

My question is: since the filtration is $\mathbb{R}$-defined, in the sense that each $\mathcal{F}^i$ is the complexification of an $\mathbb{R}$-sub vector space of $H^i_{\text{dR}}(X)$, doesn't this mean that complex conjugation acts trivially on $\mathcal{F}^qH^i_{\text{dR}}(X)$, in the sense that $\mathcal{F}^qH^i_{\text{dR}}(X) = \overline{\mathcal{F}^qH^i_{\text{dR}}(X)}$? In other words, is the complex conjugation redundant whenever my base scheme is stable under complex conjugation?