By colouring every element of $\Bbb Q$ with a unique colour and every element of $\Bbb R\setminus\Bbb Q$ with the same colour (different from all the colours used so far), we see that $\chi_\mathrm{cf}(\Bbb R)\leq\aleph_0$.
But in fact $\aleph_0$ is a lower bound in any reasonable space.
Lemma 1: Let $X$ be a $T_1$ space with at least three points. Then $\chi_{\mathrm{cf}}(X)>2$.
Proof: Suppose for a contradiction $\chi_{\mathrm{cf}}(X)=2$ as witnessed by $c\colon X\to 2$. Let $x\in X$ be such that $c(x)\neq c(x')$ for all $x'\in X\setminus\{x\}$. But now we have that $X\setminus\{x\}$ is a monochromatic open set in $X$ with at least two points, a contradiction.
Lemma 2: Let $X$ be an infinite $T_1$ space. Then $\chi_{\mathrm{cf}}(X)\geq\aleph_0$.
Proof: Suppose for a contradiction $\chi_{\mathrm{cf}}(X)=n$ as witnessed by $c\colon X\to n$. Let $x_1\in X$ be such that $c(x)\neq c(x')$ for all $x'\in X\setminus\{x_1\}$. Now $c\upharpoonright X\setminus\{x_1\}$ witnesses that $\chi_{\mathrm{cf}}(X\setminus\{x_1\})\leq n-1$. Proceed inductively to find $x_1,\ldots,x_{n-2}$ so that $c\upharpoonright X\setminus\{x_1,\ldots,x_{n-2}\}$ witnesses that the latter space has conflict-free chromatic number at most $2$, which contradicts Lemma 1.
In conclusion $\chi_{\mathrm{cf}}(\Bbb R)=\aleph_0$.