Rational functions on reduced complex varieties that extend to global holomorphic functions

The answer is yes. Ariyan Javanpeykar has contributed the hard part; here are the easy parts.

Let $\tilde{A}$ be the integral closure of $A$ in $\mathrm{Frac}(A)$ and let $\tilde{X} = \mathrm{Spec}(\tilde{A})$. Since the map $\tilde{X} \to X$ is continuous, the pull back of $h$ to $\tilde{X}$ is a continuous function. As the OP notes, this means that $h \in \tilde{A}$. So $h$ is integral over $A$.

On the other hand, Javanpeykar and Kucharczyk show that $A$ is integrally closed in the ring of holomorphic functions on $X$. We assumed that $h$ is a holomorphic function, and we have just shown that $h$ is integral over $A$. So $h \in A$.

Let $A$ be a noetherian integral domain, $K$ its field of fractions, and $f \in K$. Assume that for each maximal ideal $\frak m$ of $A$ the element $f \in K \subseteq K\otimes_{A}\hat{A}_{\frak m}$ is in $\hat{A}_{\frak m} \subseteq K\otimes_{A}\hat{A}_{\frak m}$ (here $\hat{A}_{\frak m}$ denotes the completion of $A$ at $\frak m$). Then $f \in A$.

Here is the proof. It is enough to show that $f \in A_{\frak m}$ for all maximal ideals $\frak m$; hence we can assume that $A$ is local. Set $\hat K = K \otimes_A \hat A$; then $\hat K$ contains both $K$ and $\hat A$, and the statement is that $A = K \cap\hat A \in \hat K$.

So, $f \in K \cap\hat A$. By the easy part of descent theory, it is enough to show that $f \otimes 1 = 1 \otimes f \in \hat A \otimes_A \hat A$. But $f \otimes 1 = 1 \otimes f \in \hat K \otimes_K \hat K$, because $f \in K$, and $\hat A \otimes_A \hat A$ injects into $K \otimes_A (\hat A \otimes_A \hat A) = \hat K \otimes_K \hat K$.

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