In the special case $X = \mathbb{R}$, I believe the answer is yes. I'm still interested in more general cases as well.

The proof for $X = \mathbb{R}$ goes as follows. Comments, shorter proofs or references are welcome.

Suppose that $v : \mathcal{O}(X) \to \mathbb{C}$ is a continuous complex valuation. Then we can define its real part $\mathrm{Re}(v)$ and imaginary part $\mathrm{Im}(v)$ via $\mathrm{Re}(v)(U) = \mathrm{Re}(v(U))$ and $\mathrm{Im}(v)(U) = \mathrm{Im}(v(U))$. This gives two valuations with values in $\mathbb{R}$, and we can write $v = \mathrm{Re}(v) + i\, \mathrm{Im}(v)$. So if we want to show that $v$ can be extended to a complex Borel measure, we can assume without loss of generality that $v$ takes values in $\mathbb{R}$.

We now show that there are only countably many points that have a "point mass" associated to it. Let $v : \mathcal{O}(X) \to \mathbb{R}$ be a continuous valuation, and define $w(x) = v(\mathbb{R})-v(\mathbb{R}-\{x\})$ for every real number $x$. Further, we write $W = \{ x \in \mathbb{R} : w(x) \neq 0\}$. We show that $W$ is countable. If it was not, then there would be an $N \in \mathbb{N}$ such that $W’ = \{ x \in \mathbb{R} : w(x) > 1/N \}$ is still uncountable. Uncountable subsets of $\mathbb{R}$ have a limit point, so we take a limit point $\tilde{x} \in W'$ and a sequence $(x_n)_n$ in $W’$ such that $x_n$ converge to $\tilde{x}$ and such that $W’’ = \{ x_1, x_2, x_3, \dots \} \cup \{ \tilde{x} \}$ is homeomorphic to $\{0,1,1/2,1/3,\dots\} \cup \{ 0 \}$. Now set $U_0 = \mathbb{R} - W’’$ and $U_n = U_0 \cup \{x_1, \dots, x_n\}$ for each $n>0$. Then $$U_n \cap (\mathbb{R} - \{x_n\}) = U_{n-1}$$ and $U_n \cup (\mathbb{R}-\{x_n\}) = \mathbb{R}$, so we see that $v(U_{n-1}) + v(\mathbb{R}) = v(U_n) + v(\mathbb{R}-\{x_n\})$, and as a result $v(U_n) = v(U_{n-1}) + w(x_n) \geq v(U_{n-1}) + 1/N$. But because $v$ is continuous and $U_n$ is an increasing union of open subsets, $v(U_n)$ should converge, a contradiction. So $W$ is countable.

We now show that $v$ extends to a (signed) Borel measure on $\mathbb{R}$. Define the cumulative distribution function $F(t) = v(\mathbb{R})-v((t,+\infty))$. Continuity of $v$ shows that $F$ is right continuous. We want to construct a (signed) Borel measure that has $F$ as cumulative distribution function. Since $F$ is right continuous, a Borel measure like this exists if and only if $F$ is of bounded variation (this is explained well here). Suppose that $F$ is not of bounded variation. Then we can find $a_0 < a_1 < \dots < a_n$ such that $\sum_{i = 0}^{n-1} | F(a_{i+1}) - F(a_i) | > 1$. We claim that we can assume $a_0,\dots,a_n \notin W$, for $W$ as defined above. Then because $a_i \notin W$ for each $a_i$, we find that $F(a_{i+1}) - F(a_i) = v((a_i,a_{i+1}))$ for each $i$. To make sure that $a_i \notin W$, we can replace $a_i$ by $a_i + \epsilon_i \notin W$ for $\epsilon>0$, with $\epsilon$ small enough such that still $\sum_{i = 0}^{n-1} |v((a_i+\epsilon_i,a_{i+1}+\epsilon_{i+1}))| = \sum_{i = 0}^{n-1} | F(a_{i+1}+\epsilon_{i+1}) - F(a_i+\epsilon_i) | > 1$. This is possible because $W$ is countable and $F$ is right continuous. If we now consider the disjoint open intervals $I_1^{(0)},\dots,I_{n_0}^{(0)}$ given by $(-\infty, a_0+\epsilon_0)$, $(a_0+\epsilon_0, a_1+\epsilon_1)$, …, $(a_{n-1}+\epsilon_{n-1}, a_n + \epsilon_n)$, $(a_n+\epsilon_n,+\infty)$, then we still have $\sum_{j=1}^{n_0} |v(I_j)| > 1$.

Now $F$ again fails to be of unbounded variation over at least one of the intervals $I^{(0)}_j$. We choose precisely one such interval, and we subdivide this particular interval further in the way described above, such that we end up in total with a new set of open intervals $I_1^{(1)}, \dots, I_{n_1}^{(1)}$ that again partition $\mathbb{R}$ (up to finitely many points) and such that $\sum_{j =1}^{n_1} |v(I_j)| > 2$. By repeating this process, we make in step $k$ an open partition $I_1^{(k)}, \dots, I_{n_k}^{(k)}$ of $\mathbb{R}$ (up to finitely many points) with $A(k) = \sum_{j=1}^{n_k} |v(I_j^{(k)})| > k$.

After renumbering, we can assume that the interval in step $k$ that gets subdivided in step $k+1$, is precisely the interval $I_1^{(k)}$. Further, by choosing the intervals small enough in each step, we can assume that $v(I_1^{(k)})$ converges to a constant $C \in \mathbb{R}$ as $k$ goes to infinity. Now let $\Omega$ be the set of intervals that do not have a further subdivision, i.e. the intervals of the form $I_j^{(k)}$ for some $k$ and $j\geq 2$. Then $\sum_{I \in \Omega} v(I)$ converges unconditionally to $v(U)$ with $U = \bigsqcup_{I \in \Omega} I$. But then the series is also absolutely convergent by Riemann's Rearrangement Theorem, so $A = \sum_{I \in \Omega} |v(I)| < +\infty$. We now see that $k < A(k) = \sum_{j=1}^{n_k} |v(I_j^{(k)})| \leq A + |v(I_1^{(k)})|$. Because $|v(I_1^{(k)})|$ converges to $C$ for $k \to +\infty$, this gives a contradiction.

Because $F$ is right continuous and of bounded variation, there is a continuous Borel measure $\mu$ such that $F(t) = \mu((-\infty, t])$. It remains to show that $\mu$ restricts to $v$ on open subsets. First one can show that $\mu(\mathbb{R}) = v(\mathbb{R}) = \lim_{x\to+\infty} F(x)$. Then from the definition of $F$ it follows that $\mu((a,+\infty))=v((a,+\infty))$ for all $a \in \mathbb{R}$. Further, $$\begin{align*}\mu((-\infty,b)) &= \lim_{\epsilon \to 0} \mu((-\infty,b-\epsilon]) = \lim_{\epsilon \to 0} F(b-\epsilon) = \lim_{\epsilon \to 0} v(\mathbb{R}) - v((b-\epsilon,+\infty)) \\ &= \lim_{\epsilon \to 0} v((-\infty,b-\epsilon)) = v((-\infty,b))\end{align*},$$ where in the second last equality we use that we can take the limit over those $\epsilon > 0$ with $b-\epsilon \notin W$. Since $\mu$ and $v$ agree on the intervals $(a,+\infty)$ and $(-\infty,b)$, they agree on all open sets.