I think it is not possible. Such a ring isomorphism $\phi$ should also preserve the order structure, because in both rings non-negative elements are exactly the squares; as a consequence, it must also preserve the constant functions, since it preserves the constant $1$. In other words, $\phi$ is an ordered $\mathbb{R}$-algebras isomorphism.
In both rings, characteristic functions of singletons can be characterized in terms of the ordered $\mathbb{R}$-algebra structure, as e.g. those idempotents $u$ such that any positive element smaller than $u$ is a scalar multiple of $u$ ( that is "$0\le f\le u$ implies $f=\lambda u$ ").
Note that the ring $S$ contains all characteristic functions of singletons of $\mathbb{R}$.
Since $\phi$ preserves the ordered $\mathbb{R}$-algebras structure, if $u:=\chi_{\{t\}}$ is a characteristic function of a singleton of $\mathbb{R}$, then $\phi(u)$ is also a characteristic function of a singleton $\chi_{\{x\}}$ of $X$, necessarily open because $\chi_{\{x\}}\in C(X)$.
This way we have defined an injective map $\psi:\mathbb{R}\to X$ such that for all $t\in\mathbb{R}$ one has $\phi(\chi_{\{t\}})= \chi_{\{\psi(t)\}}\, .$ Since $\psi(\mathbb{R})$ is an open subset of $X$ with the discrete topology, $C(X)$ also contains the characteristic function of $\psi(\mathbb{Q})$. Again by the order properties of $\phi$, this allows to conclude that $S$ contains the characteristic function of $\mathbb{Q}$, a contradiction.