Let R be a noetherian normal domain (if it makes any difference, I'm happy to assume R is also local).

If $p$ is a height one prime, then the localization $R_p$ is a dvr, hence the maximal ideal $pR_p$ is principal, generated by a uniformizer $\pi$.

It's easy to see we can pick $\pi \in p$. Similarly, if $\pi' \in p$ is another uniformizer then there exist $s,t \in R - p$ such that $t\pi' = s\pi$.

Can one do better than this?

More precisely, is there a uniformizer $\pi_0 \in p$ such that any other uniformizer $\pi'$ is of the form $\pi' = s\pi_0$ for $s \in R - p$?

EDIT: David's answer shows that an affirmative answer to the question above actually implies p itself is principal (which is very strong). So here's a weaker question.

Since p is noetherian, it can be generated by finitely many elements $f_1,\ldots,f_r$. Can one arrange so that $ord_p(f_i) < ord_p(f_{i+1})$?

The motivation for this question comes from the following standard example. Take $R = k[x,y,z]/(z^2 - xy)$. Take $p = (x,z)$. This is a height one prime ideal. Here $z$ is a uniformizer. Although $p$ is not principal, one needs only one element of order one to generate it.

EDIT 2: I would also happily assume that R is moreover complete and an algebra over a field (of say characteristic zero).