Is there a model of $\operatorname{Th}(\mathbb{R})$ which is not a complete ordered field?

Actually $\mathbb{R}$ is the only complete ordered field, see this MO thread; basically it's because order completeness implies that the order must be archimedean, and any archimedean ordered field is a subfield of $\mathbb{R}$. $Th(\mathbb{R})$ is just the theory of real closed field, so any real closed field other than $\mathbb{R}$ would be an example: real algebraic numbers, real closure of $\mathbb{R}(x)$, etc.

Note that there is another notion of Cauchy completeness of an ordered field. There are plenty of examples of Cauchy complete ordered field.

Of course!

Consider the hyperreal numbers ${}^* \mathbb{R}$ for instance. Then the set $\mathbb{R} \subseteq {}^* \mathbb{R}$ is bounded (by any unbounded natural, say) yet it has no least upper bound. Since the hyperreals are constructed to satisfy $\mathsf{Th}(\mathbb{R})$, we get the claim.

I hope this helps ^_^

Every ultrapower of $\mathbb{R}$ is elementarily equivalent to $\mathbb{R}$ (i.e. it has the same first-order theory) but it need not be complete. In particular, an ultrapower of $\mathbb{R}$ with respect to a non-principal ultrafilter on $\omega$ will not be complete.

Proof. Let $\mathbb{P}$ be such an ultrapower of $\mathbb{R}$. In other words, elements of $\mathbb{P}$ are functions $f\colon \omega \to \mathbb{R}$ modulo the equivalence relation determined by the ultrafilter. We identify $\mathbb{R}$ with the substructure of $\mathbb{P}$ which consists of constant functions (up to equivalence in the ultrafilter). There is some $f \in \mathbb{P}$ such that $0 < f < r$ in $\mathbb{P}$ for each $r \in \mathbb{R}^{+}$, e.g. $f(n) := \frac{1}{n}$. If $\mathbb{P}$ were complete, then the set of all such $f \in \mathbb{P}$ would have a supremum, say $s$. But then $s < 2s$, so there is some $r \in \mathbb{R}^{+}$ such that $0 < r < 2s$. This implies that $0 < \frac{r}{2} < s$, contradicting the assumption that there is no real strictly between $0$ and $s$.