This is true when $R$ and $\tilde R$ are both Noetherian, e.g. when $R$ is Noetherian and Japanese. It might be possible to weaken some of these hypotheses.
Proposition. Let $R$ be a domain such that $R$ and $\tilde R$ are Noetherian. If $p \in R$ is a prime element, then $p$ is a prime element in $\tilde R$.
Proof. By assumption, $\mathfrak p = (p)$ is a prime ideal. By Krull's Hauptidealsatz, this implies that $\mathfrak p$ has height $1$, hence $R_\mathfrak p$ is a $1$-dimensional domain. Since its maximal ideal $\mathfrak pR_\mathfrak p$ is principal, we conclude that $R_\mathfrak p$ is a DVR [AM, Prop. 9.2] with uniformiser $p$; in particular $R_\mathfrak p$ is normal.
On the other hand, normalisation commutes with localisation [AM, Prop. 5.12]. Thus, $$(\tilde R)_\mathfrak p = (R_\mathfrak p)^\sim = R_\mathfrak p,$$ since $R_\mathfrak p$ is normal. That is, the natural map $R \to \tilde R$ becomes an isomorphism when tensoring with $R_\mathfrak p$, hence also when tensoring with $\kappa(\mathfrak p) = R_\mathfrak p/\mathfrak pR_\mathfrak p$. The primes of $\tilde R \otimes_R \kappa(\mathfrak p)$ are the primes of $\tilde R$ lying over $\mathfrak p$ [AM, Exc. 3.21(iv)], so we conclude that there is a unique such prime $\mathfrak q$. We clearly have $(p) = \mathfrak p\tilde R \subseteq \mathfrak q$; we want to show that this is an equality.
Consider the map $R/\mathfrak p \to \tilde R/\mathfrak p\tilde R$, and note that $\mathfrak q/\mathfrak p\tilde R \subseteq \tilde R/\mathfrak p\tilde R$ is a minimal prime. If $\mathfrak q' \subseteq \tilde R$ is a different prime that is minimal over $\mathfrak p\tilde R = (p)$, then by Krull's hauptidealsatz $\mathfrak q'$ has height $1$ in $\tilde R$. If $\mathfrak p' = \mathfrak q' \cap R$, then going up for $R_{\mathfrak p'} \to (\tilde R)_{\mathfrak p'}$ shows that $\mathfrak p'$ has height $1$ in $R$. Since $\mathfrak p \subseteq \mathfrak p'$, we conclude that $\mathfrak p = \mathfrak p'$ for height reasons, hence $\mathfrak q = \mathfrak q'$.
Thus, $\tilde R/\mathfrak p\tilde R$ has a unique minimal prime $\mathfrak q/\mathfrak p\tilde R$. Hence, $\mathfrak q/\mathfrak pS$ is the nilradical of $\tilde R/\mathfrak p\tilde R$, so there exists an $r \in \mathbb Z_{> 0}$ such that $\mathfrak q^r = \mathfrak pS$. But $p$ is a uniformiser of the DVR $\tilde R_\mathfrak q \cong R_\mathfrak p$, so we must have $r = 1$, i.e. $\mathfrak q = \mathfrak pS$. $\square$
Remark. In geometric language, we proved:
- There is a unique irreducible divisor $V(\mathfrak q) \subseteq \operatorname{Spec} \tilde R$ dominating the irreducible divisor $V(\mathfrak p) \subseteq \operatorname{Spec} R$;
- The locus $V(p) \subseteq \operatorname{Spec} \tilde R$ does not split off a new component of higher codimension;
- The uniformiser $p$ for the divisor $V(\mathfrak p) \subseteq \operatorname{Spec} R$ remains a uniformiser for $V(\mathfrak q) \subseteq \operatorname{Spec} \tilde R$ (there is no ramification).
References.
[AM] Atiyah, M.F.; Macdonald, I.G., Introduction to commutative algebra. Addison-Wesley Publishing Company (1969). ZBL0175.03601.