uniqueness of uniformizers 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).
 A: 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.

Response to the new question: Let $p_k = \{ x \in R : \mathrm{ord}_p(x) \geq k \}$. So $p = p_1$, but note that $p_2$ is probably larger than $p^2$. (In your example, $y \in p_2$ but $y \not \in p^2$.) I claim your new condition is equivalent to saying that $p/p_2$ is a free $R/p$ module. Proof: If $p = \langle f_1, f_2, \ldots, f_r \rangle$ as you propose, then $p/p_2 = \langle f_1 \rangle$. Conversely, if $p/p_2 = (R/p) f$, then lift $f$ to $f_1 \in p$ and $p/(R f_1 + p_2)=0$. Choosing generators for $p_2$, we have your claim.
I don't know a general result about when $p/p_2$ is free, but I see three observations: (1) It is not always true. Take $R = k[w,x,y,z]/(wz-xy)$ and $p = \langle w,x \rangle$, so $R/p = k[y,z]$. Then $p_2 = \langle w^2, wx, x^2 \rangle$ and $p/p_2$ is the $k[y,z]$ module on generators $w$ and $x$ subject to the relation $z \cdot w = y \cdot x$, which is not true. (2) It is true in Dedekind domains, since $p_2 = p^2$ and $p/p^2$ is a field. (3) It is true if $R$ is normal and local of dimension $2$, since then $R/p$ is a dvr. This argument was false. I'll modify it to (3') The statement is true if $R$ is normal and $R/p$ is a dvr.  Over a dvr, every torsion free finitely generated module is free. The $R/p$ module $p/p_2$ is torsion free, and is rank $1$ since $R$ is regular at $p$.
