This is the same as asking that $p$ is principal. In one direction, if $p = (\pi)$, then take $\pi$ to be the uniformizer.
In the reverse direction, suppose $\pi$ has the stated property. I claim that $p = (\pi)$. Let $f$ be a nonzero element of $p$; we must show that $f$ is a multiple of $\pi$. Since $R$ is a noetherian domain, $\bigcap p^n = (0)$, so there is some $n$ for which $f \in p^n \setminus p^{n+1}$. Thus, $f = \pi_1 \pi_2 \cdots \pi_n$ where each $\pi_i \in p \setminus p^2$, and is thus a uniformizer. By the hypothesis, each $\pi_i$ is of the form $s_i \pi$, so $f = (\prod s_i) \pi^n$. We have shown that $\pi$ divides $f$.
A Krull domain has all height one primes principal iff it is a UFD. A normal noetherian domain is Krull. Thus, your condition holds for all height one primes if and only if $R$ is a UFD.