This is one of those questions that feels true, but I can't prove or disprove it.
Let $K = \mathbb{Q}(\sqrt{D})$, where $D < 0$ is squarefree, and let $k>0$ be some positive integer. If you take some $a = a_0 + a_1p^k\sqrt{D}\in \mathbb{Z}_{(p)}[p^k\sqrt{D}]$ such that $$N_{K/\mathbb{Q}}(a) = p^\alpha u, \alpha >0$$ where one of $v_p(a_0)$ or $v_p(a_1) = 0$, and also $v_p(u) =0$. If there is a $b\in K$ such that $ab = p^\beta$, with $\beta < \alpha$, then must it be the case that $b\not\in \mathbb{Z}_{(p)}[p^k\sqrt{D}]$.
Here I mean fractions without $p$'s in the denominator when I write $\mathbb{Z}_{(p)}$.
This is equivalent (I think) to the statement of
The minimal power of $p$ in the ideal $(a) = (a_0+a_1p^k\sqrt{D})\subseteq \mathbb{Z}_{(p)}[p^k\sqrt{D}]$ where $v_p(a_0)$ or $v_p(a_1) = 0$ is $p^{v_p(N_{K/\mathbb{Q}}(a))}$.
The main problem is that the ring in question is just shy of being a DVR, and so valuation theory goes out the window; so, it's entirely possible that this is not the case, which would be a shame, but I'm hopeful! :)
Any and all help is appreciated! Thanks in advance!
It is, in fact, true!
Here's a quick sketch of why:
Suppose $a,b\in \mathbb{Z}_{(p)}[p^k\sqrt{D}]$ with $N_{K/\mathbb{Q}}(a) = p^\alpha u$ and $u\in \mathbb{Q}$ such that $v_p(u) = 0$. Say that $$ab = (a_0+a_1p^k\sqrt{D})(b_0+b_1p^k\sqrt{D}) = p^\beta$$ is minimal in $\beta$. Then, $a_0b_1p^k + b_0a_1p^k$ must be zero (the coefficient on the $\sqrt{D}$ term).
This is (the negative of) the determinant of the matrix $$A = \begin{bmatrix}a_0 & a_1p^k\\ b_0 & -b_1p^k\end{bmatrix}.$$
So, we need this matrix to have rank 1 over $\mathbb{Q}$; in particular, this shows that the conjugate of $b$ is a $\mathbb{Z}_{(p)}$ multiple of a, and so the product $ab$ is a $\mathbb{Z}_{(p)}$ unit times the norm of $a$, proving the desired result.