I've made some progress on the following problem:
Let $R$ be a Noetherian ring, $R \subseteq S$ an extension of rings, and $x \in S$. Show that $x$ is integral over $R$ if for every minimal prime $\mathfrak q$ of $S$, $\overline{x} \in S/\mathfrak q$ is integral over $R/ \mathfrak q \cap R$.
I believe I've solved the problem under the hypothesis that $S$ is Noetherian. In this case, there are only finitely many minimal prime ideals $\mathfrak q_1, ... , \mathfrak q_s$ of $S$. By hypothesis, for each $i$, there exists a monic polynomial $f_i(X) \in R[X]$ such that $f_i(x) \in \mathfrak q_i$. So then $$f_1(x) \cdots f_t(x) \in \mathfrak q_1 \cap \cdots \cap \mathfrak q_t = \sqrt{0}$$ so $f_1(x)^m \cdots f_t(x)^m = 0$ for some $m \geq 1$. But then $x$ is a root of the polynomial $f_1(X)^m \cdots f_t(X)^m$, which is monic, so $x$ is integral over $R$.
If $S$ is not Noetherian, I can't expect $S$ to have finitely many minimal prime ideals. I haven't figured out how to use the hypothesis that $R$ is Noetherian yet. Is there some way I can adapt the above argument?
We have the tower of rings $R \subseteq R\left[x\right] \subseteq S$. For every minimal prime ideal $\mathfrak{p}$ of $R\left[x\right]$, there exists a minimal prime ideal $\mathfrak{q}$ of $S$ lying over $\mathfrak{p}$ (this is true for any ring extension, as Youngsu proves in the comments).
Being a finitely generated $R$-algebra, $R\left[x\right]$ is Noetherian. Hence, $R\left[x\right]$ has only finitely many minimal prime ideals, say $\mathfrak{p}_{1}, \ldots, \mathfrak{p}_{t}$. Let $\mathfrak{q}_{1}, \ldots, \mathfrak{q}_{t}$ be minimal prime ideals of $S$ with $\mathfrak{q}_{i} \cap R\left[x\right] = \mathfrak{p}_{i}$ for each $i$.
As you state, for each $i$, there exists a monic polynomial $f_{i}\left(X\right) \in R\left[X\right]$ with $f_{i}\left(x\right) \in \mathfrak{q}_{i}$. Since $x \in R\left[x\right]$ and all the coefficients of $f_{i}\left(X\right)$ are in $R \subseteq R\left[x\right]$, it follows that $f_{i}\left(x\right) \in R\left[x\right]$. Thus, $f_{i}\left(x\right) \in \mathfrak{q}_{i} \cap R\left[x\right] = \mathfrak{p}_{i}$.
From here, you can proceed exactly as you suggest in your question, using that $f_{1}\left(x\right)\cdots f_{t}\left(x\right) \in \sqrt{0_{R\left[x\right]}}$.