The answer is no, essentially because $\mathbb{R}$ embeds as a locally compact open subspace of $\beta\mathbb{R}$, and $\mathbb{R}$ is not an F-space.

In detail, for the purposes of this answer I will write $\mathbb{R} \subseteq \beta\mathbb{R}$. The facts we will use are that $\mathbb{R}$ is an open subspace of $\beta\mathbb{R}$ because it is locally compact, and that compact subsets of $\mathbb{R}$ are compact, and therefore closed, in $\beta\mathbb{R}$.

Consider $(0,1)$ and $(1,2)$ in $\mathbb{R}$. These are disjoint opens in $\mathbb{R}$, therefore in $\beta\mathbb{R}$ (because $\mathbb{R}$ is an open subset). The first set $(0,1) = \bigcup\limits_{i=1}^\infty [2^{-i},1-2^{-i}]$, so is $F_\sigma$ in $\beta\mathbb{R}$ (because closed bounded intervals are compact in $\mathbb{R}$, and therefore closed in $\beta\mathbb{R}$). A similar argument shows that $(1,2)$ is $F_\sigma$. Their closures are $[0,1]$ and $[1,2]$ in $\mathbb{R}$, and as these are compact, they are also their closures in $\beta\mathbb{R}$. These are not disjoint.

Some set-theory, Stone–Čech, and F-spacesdoi.org/10.1016/j.topol.2011.06.007 and some other related papers cited there? $\endgroup$ – Martin Sleziak Dec 4 '18 at 9:10Rings of continuous functions in which every finitely generated ideal is principaldoi.org/10.1090/S0002-9947-1956-0078980-4 seems like a reasonable candidate - and I suppose that after that we can delete all comments related to clarification of the question. $\endgroup$ – Martin Sleziak Dec 4 '18 at 9:161more comment