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.