Let $E$ be a complex vector space endowed with a Hermitian metric, it can be seen as the complexification of a real vector space $E_R$, with its metric coming from an inner product on $E_R$. Let $X$ be a partial flag variety over $E$ of some fixed dimensions $d_1< \cdots < d_s$.
The subset of flags that are complexifications of real flags is Zarisky dense in $X$.
$\def\CC{\mathbb{C}}\def\RR{\mathbb{R}}$This yields to a standard lemma: Let $X$ (in your situation, the flag manifold) be a smooth connected complex variety equipped with an anti-holomorphic involution $\sigma$ (in your case, replacing a flag by its complex conjugate) and let $X_{\RR}$ be the fixed points of $\sigma$. If $X_{\RR}$ is nonempty, then it is Zariski dense in $X$.
Proof: Let $Y \subseteq X$ be a Zariski closed set containing $X_{\mathbb{R}}$. For any point $x \in X_{\mathbb{R}}$, if we pass to a small enough neighborhood $U$ of $x$, then $Y\cap U$ will be cut out of $U$ by the vanishing of algebraic functions. Pass to a small analytic disc $U$ around some $x \in X$. In this disc, $X_{\RR}$ looks like an open ball of $\RR^{n}$ sitting in an open ball of $\CC^n$. But a holomorphic function vanishing on the intersection of an open ball with $\mathbb{R}^n$ is $0$, so the only function on $U$ which vanishes on $X_{\RR} \cap U$ is $0$. Thus, $Y \cap U = U$. We have shown that $Y$ is open in $X$, and it is also closed. By hypothesis, $X$ is connected and $Y$ is nonempty, so $X=Y$. $\square$
Example to show you need $X_{\RR}$ nonempty: $x^2+y^2=-1$ is smooth but has no real points.
Example to show you need $X$ smooth: Let $X = \{ x^2+y^2+z^2=0 \}$, the Zariski closure of $X_{\RR}$ is $\{ x=y=z=0 \}$.
