Let $(R,+,\cdot ,\Phi)$ be a euclidean domain where $\Phi$ denotes the Euclidean function, that is a function $R \setminus \lbrace 0 \rbrace \rightarrow \mathbb N$ such that :
$\forall a \in R \quad \forall b \in R \setminus \lbrace 0 \rbrace \quad \exists q,r \quad | \quad a=qb+r \quad \text{and} \quad (r=0 \quad \text{or} \quad \Phi(r) < \Phi(b))$
$\forall a \in R \quad \forall b \in R \setminus \lbrace 0 \rbrace \quad \Phi(a) \leq \Phi(ab)$
Define a strict partial order on $R$ by
$a < b \iff \Phi(a) < \Phi(b)$.
I strongly suspect that this order must have some relation with the order induced by divisibility. I wouldn't say that it's the same because I don't see why $\Phi(a) < \Phi(b)$ should imlply that $a|b$ but is there anything remarkable about these two partial orders ?
The answer is: not much. Consider the degree function $D$ over $\mathbb{Q}[X]$, which is an Euclidean function. Then $D(f)\leq D(g)$ does not imply anything about the divisibility of $f$ and $g$.
In fact, even in $\mathbb{Z}$ with the Euclidean function $|\cdot|$ you cannot infer any divisibility property from $|n|\leq|m|$. That's precisely why Euclidean domains are of some value, because the information we gain through the Euclidean function is not superfluous.