Is there a subfield of $\mathbb{R}$ that is a proper elementary extension of $\mathbb{Q}$?

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 $\mathbb{R}$.