Let $K$ be a field, $K_s$ its separable closure, $K$ $\subseteq$ $F$ $\subseteq$ $K_s$ an extension with $[F:K]$ $=$ $n$, $R$ $\subseteq$ $K$ a Dedekind domain with quotient field $K$, $S$ the integral closure of $R$ in $F$, and $\mathfrak{p}$ a maximal ideal of $R$.

Can we find a $d>n$ with $(d,n)=1$ and an extension $K$ $\subseteq$ $\widetilde{K}$ $\subseteq$ $K_s$ with $[\widetilde{K}:K]$ $=$ $d$ such that $\widetilde{K}$ and $F$ are linearly disjoint over $K$ (in $K_s$) and, if $\widetilde{R}$ denotes the integral closure of $R$ in $\widetilde{K}$, the integral closure of $R$ in $\widetilde{K}F$ is equal to $\widetilde{R}S$, and $\mathfrak{p}$ is inert in $\widetilde{K}$ (i.e. $\mathfrak{p}\widetilde{R}$ is a prime ideal of $\widetilde{R}$) ?

It is okay to assume that $K$ and $F$ are algebraic number fields, or even that $R$ $=$ $\mathbb{Z}$.

In the case where $K$ and $F$ are algebraic number fields, if $d$ $>$ $n$ is a prime number, any extension $K$ $\subseteq$ $\widetilde{K}$ of degree $d$ is linearly disjoint to $F$. And when the discriminants of $S$ and $\widetilde{R}$ over $R$ are relatively prime in $R$, the integral closure of $R$ in $\widetilde{K}F$ equals $\widetilde{R}S$. (Cf. Fröhlich and Taylor, Algebraic Number Theory, Ch. III, 2.13). If $\mathfrak{p}\cap\mathbb{Z}$ $=$ $p\mathbb{Z}$, by Zsigmondy's theorem we can find a prime number $q$ such that $p$ has order $d$ in $\mathbb{F}_q^{\times}$. Then $p$ splits as a product of $(q-1)/d$ primes of degree $d$ in $\mathbb{Q}(\zeta_q)$. So if $d$ and $(q-1)/d$ are relatively prime, $p$ is inert in the unique subfield $D$ of $\mathbb{Q}(\zeta_q)$ of degree $d$ over $\mathbb{Q}$. Starting out with a prime $d$ that is larger than the residue class degree of $\mathfrak{p}$ over $p$ as well, $\mathfrak{p}$ will be inert in the compositum $\widetilde{K}$ $=$ $KD$. And taking $d$ also larger than the absolute value of the discriminant of $F$ over $\mathbb{Q}$, $q$ is relatively prime to the discriminant of $S$ over $R$, since $q$ $>$ $d$. So this field $\widetilde{K}$ will do the trick.

However, there is no guarantee that $(d,(q-1)/d)$ $=$ $1$, and I would be much obliged to learn how such a construction could be made to work (without appealing to extended Riemann hypotheses).

The existence of such a $\widetilde{K}$ in the general (Dedekind domain) setup is used in this Journal of Algebra paper by Ilaria Del Corso and Roberto Dvornicich. In the paragraph following their Lemma 4, the authors optimistically "let $\widetilde{K}$ be an unramified extension of $K$ of degree $d$ such that $\mathfrak{p}$ is inert in $\widetilde{K}$" (with any $d$ that is large enough and relatively prime to $n$). They then show (Lemma 5) that $\widetilde{K}$ and $F$ are linearly disjoint and that the integral closure of $R$ in $\widetilde{K}F$ is $\widetilde{R}S$, using the fact that $\widetilde{K}$ is unramified over $K$. But, aside from $\mathbb{Q}$, various other fields $K$ exist that do not admit any non-trivial unramified extensions whatsoever (see this answer).