The issue here is what conditions one requires on a valuation function. From https://en.wikipedia.org/wiki/Euclidean_domain:
Let $R$ be an integral domain. A Euclidean function [or valuation] on
$R$ is a function $f$ from $R \setminus \{0\}$ to the non-negative
integers satisfying the following fundamental division-with-remainder
property:
- $(EF_1)$ If $a$ and $b$ are in $R$ and $b$ is nonzero, then there are $q$ and $r$ in $R$ such that $a = bq + r$ and either $r = 0$ or $f(r) < f(b)$.
A Euclidean domain is an integral domain which can be endowed with at
least one Euclidean function. It is important to note that a
particular Euclidean function $f$ is not part of the structure of a
Euclidean domain: in general, a Euclidean domain will admit many
different Euclidean functions.
Most algebra texts require a Euclidean function to have the following
additional property:
- $(EF_2)$ For all nonzero $a$ and $b$ in $R$, $f(a) ≤ f(ab)$.
Let me interrupt the Wikipedia quotation here to point out that property $EF_2$ is essentially what you are asking about. If $EF_2$ holds for some valuation $\nu$, then of course $a \mid b \implies \nu(a) \le \nu(b)$.
So one way to paraphrase your question is: What if we are using a definition of valuation that requires $EF_1$ but not $EF_2$?
Now we return to our quote from Wikipedia:
However, one can show that $EF_2$ is superfluous in the following
sense: any domain $R$ which can be endowed with a function $g$ satisfying
$EF_1$ can also be endowed with a function $f$ satisfying $EF_1$ and
$EF_2$: indeed, for $a \in R \setminus \{0\}$ one can define
$f(a)$ as follows:
$$f(a) = \min_{x \in R \setminus \{0\}} g(xa)$$
In words, one may define $f(a)$ to be the minimum value attained by $g$ on
the set of all non-zero elements of the principal ideal generated by
$a$.
So one way to answer your question is: Let's suppose you are working with a valuation $\nu$ that does not satisfy $EF_2$. By the result above, you can switch to another one that does. With respect to that new valuation, the categorical definition of GCD coincides with the "old" definition.