Mostly just the question in the title. This question comes out of the fact that the order relation is definable in both $\mathbb{R}$ and $\mathbb{Q}$, but the definitions are very different: In $\mathbb{R}$, $a \leq b$ if and only if $\exists c(c^2 = b-a)$. On the other hand, in $\mathbb{Q}$, $\mathbb{Z}$ is a definable subset (via some very non-obvious algebra black magic) and $\mathbb{N}$ is a definable subset of $\mathbb{Z}$ (via the four square theorem), so the positive rationals can be defined as those which are ratios of strictly positive integers. So you can say $a\leq b$ if and only if $b-a$ is positive or $0$.
Since $\mathbb{Q}$ is bi-interpretable with $\mathbb{N}$, every proper elementary extension of $\mathbb{Q}$ contains non-standard integers and is therefore non-Archimedean as an ordered field in terms of the definition of order in $\mathbb{Q}$, but since this is different from the definition of order it's possible that some elementary extension of $\mathbb{Q}$ can be embedded in $\mathbb{R}$ as a subfield.
The comment by orangeskid is spot on. Since $\leq$ is definable by an existential formula over $\mathbb{Q}$, and the same formula suffices to define $\leq$ on $\mathbb{R}$, every embedding from a model of $\mathrm{Th}(\mathbb{Q})$ to a model of $\mathrm{Th}(\mathbb{R})$ must be order-preserving.
Explicitly, suppose $f\colon Q\to R$ is an embedding of fields, where $Q\models \mathrm{Th}(\mathbb{Q})$ and $R\models \mathrm{Th}(\mathbb{Q})$. Suppose $a\leq b$ in $Q$. By the four square theorem in $\mathbb{Q}$, there exist $c_1,c_2,c_3,c_4\in Q$ such that $c_1^2+c_2^2+c_3^2+c_4^2 = b-a$. So $f(c_1)^2+f(c_2)^2+f(c_3)^2+f(c_4)^2 = f(b)-f(a)$. And a sum of squares is always positive in $\mathbb{R}$, so $f(a)\leq f(b)$ in $R$.
As you note, since any proper elementary extension of $\mathbb{Q}$ contains a nonstandard integer ($N$ such that $n\leq N$ for all $n\in \mathbb{N}$), no proper elementary extension of $\mathbb{Q}$ embeds in $\mathbb{R}$.