Let us first observe that an ideal $I$ of a commutative ring $R$ with identity is *minimal* in OP's sense if and only if $I$ is a non-zero simple module over $R$. From now on, we will favour this terminology.

Note also that a topological space $X$ with trivial topology, i.e., the open sets of $X$ are $X$ and $\emptyset$, is completely regular for a trivial reason.
Let $C(X)$ be the ring of continuous real-valued functions on $X$. If $X$ is a topological space with trivial topology, then the ring $C(X)$ is isomorphic to $\mathbb{R}$ and hence is a simple ring.

This leads to

**Claim.** Let $X$ be a completely regular topological space.
Then $I \subseteq C(X)$ is a non-zero simple ideal if and only if $I$ is the set of continuous functions $f$ such that either $f = 0$ or $\{ x \in X \,|\, f(x) \neq 0 \} = Y$ where $Y = Y(I) \subseteq X$ is an open subspace depending only on $I$ and whose induced topology is trivial. In particular, any non-zero simple ideal of $C(X)$, if it exists, is isomorphic to $\mathbb{R}$.

If $X$ is a completely regular topological space and $Y$ is an open subspace of $X$ with trivial induced topology, then $Y$ is also closed.
In particular, if $x \in X$ is an isolated point, then $\{x\}$ is closed and the ideal consisting of the continuous functions $f$ such that $f(y) = 0$ for every $y \neq x$ is simple. I am indebted to **Zach Teitler** who mentioned this fact first.

*Proof*.
Let $R = C(X)$ and let $Y \subseteq X$ be a topological subspace whose induced topology is trivial. Then any $f \in R$ is constant on $Y$ and since $Y$ is both closed and open, any continuous function $f$ defined on $Y$ can be continuously extended to $X$ by setting $f(x) = 0$ for $x \notin Y$. Thus the ideal consisting of all functions which are zero on $X \setminus Y$ is a simple ideal.
Consider now a simple ideal $I = Rf$ with $f \neq 0$ and let $Y = \{ x \in X \,|\, f(x) \neq 0 \}$. Since $Rf = Rf^2$, the function $1/f$ defined over $Y$ can be continuously extended to $X$. From this we infer that $Y$ is closed. Indeed, if there is $x \in \overline{Y} \setminus Y$, then the continuous extension of $1/f$ cannot be bounded on an open subset containing $x$, a contradiction. Now let $g \in R$ be such that its restriction to $Y$ is not zero. Since $Rfg = Rf = I$, we deduce that $g$ has no zero in $Y$. Let us prove that $g$ is constant. Reasonning by contradiction, we consider $x,y \in Y$ such that $g(x) \neq g(y)$. Let $F$ be the pre-image under $g$ of a closed segment in $\mathbb{R}$ that contains $g(x)$ but not $g(y)$. Then $F$ is a closed set that contains $x$ but not $y$. Since $X$ is completely regular, we can find a function $h \in R$ such that $h$ vanishes on $F$ and $h(y) = 1$. This is impossible since we have established before that $h$ cannot have any zero in $Y$. Hence every $g \in R$ is constant over $Y$. Since $Y$ is both closed and open, every continuous function on $Y$ extends continuously to $X$. Therefore any continuous function on $Y$ is constant. As $Y$ is also a completely regular topological space, its topology is trivial.