The answer to your question depends critically on what you mean by a "complete ordered field" $(F,<)$. Here are two rival definitions:
[added: sequentially] Cauchy complete: every Cauchy sequence in $F$ converges.
Dedekind complete: every nonempty subset $S \subset F$ which is bounded above has a least upper bound.
(There are in fact many other axioms equivalent to 2): that every bounded monotone sequence converges, that $F$ is connected in the order topology, the principle of ordered induction holds, and so forth.)
It turns out that there is a unique Dedekind complete ordered field up to (unique!) isomorphism, namely the real numbers $\mathbb{R}$. Famously $\mathbb{R}$ is also Cauchy complete -- or, if you like, Dedekind complete ordered fields satisfy the Bolzano-Weierstass theorem, which is enough to make Cauchy sequences converge -- so that Dedekind completeness implies Cauchy completeness.
The converse is true with an additional hypothesis: an Archimedean Cauchy-complete field is Dedekind complete. I show this in $\S 12.7$ of these notes using somewhat more sophisticated methods (namely Cauchy nets). For a more elementary proof, see e.g. Theorem 3.11 of this nice undergraduate thesis.
On the other hand, just as one can take the "Cauchy" completion of any metric space (or normed field) and get a complete metric space (or complete normed field), one can take the Cauchy completion of a non-Archimedean ordered field and get an ordered field which is Cauchy complete but not Dedekind complete. The easiest example of such a field is probably the
rational function field $\mathbb{R}(t)$ with the unique ordering that makes $t$ positive and infinitely large.
For some reason these subtleties seem to be hard to find in standard analysis texts. I myself didn't learn about them until rather recently (so, several years after my PhD). I actually wrote up some of this material as supplemental notes for a sophomore-junior level course I am currently teaching on sequences and series...but I have not as yet been able to make myself inflict these notes on my students. I talked about ordered fields in several lectures and it seemed to be one level of abstraction beyond what they could even meaningfully grapple with (so it started to seem a bit pointless).
The standard counterexample is the field $F$ of finitely-tailed Laurent series over $\mathbb{R}$: series of the form $\sum a_n t^n$ where $a_n \in \mathbb{R}$, exponents $n \in \mathbb{Z}$, but with only finitely many negative exponents (i.e., $a_n = 0$ for all but finitely many negative integers $n$.)
Addition and multiplication are defined just like for power series; you should verify that the "finitely many negative exponents" condition is essential to ensuring that multiplication of finitely-tailed Laurent series is well-defined. Check that non-zero elements of $F$ have multiplicative inverses: a series that starts with $a_n t^n$ has an inverse that starts with $a_n^{-1} t^{-n}$. (Here, and below, when I say that a nonzero element of $F$ starts with $a_n t^n$, I mean that $n$ is the lowest integer for which $a_n \ne 0$.)
Ordering is defined as follows: An element is positive if its starting term has positive coefficient. It's easy to check: (1) for every nonzero $x \in F$, exactly one of $x$ and $-x$ is positive; (2) the sum and product of 2 positive elements is positive.
So $F$ is an ordered field. Note that $F$ is not Archimedean: $t^{-1}$ is bigger than every integer. The field $F$ also does not satisfy the least upper bound property: $t^{-1}$ is an upper bound for the integers, but there is no least upper bound (check that if $x$ is such an upper bound, then $x/2$ is a smaller one.)
To discuss Cauchy completeness, we have to put a metric on $F$, as follows. Pick any positive real number $q>1$. If a nonzero element $x \in F$ starts with $a_n t^n$, then define $|x| = q^{-n}$ (and of course define $|0|=0$), and then define the distance between $x$ and $y$ to be $|x-y|$. Check that this is a metric space. (Caution: This doesn't extend the existing metric on $\mathbb{R}$; all real numbers have the same size under this metric.) In fact, it makes $F$ into a valued field; we have $|xy| = |x||y|$.
This is a "non-Archimedean" metric because it satisfies the strong triangle inequality $d(x,z) \le \max(d(x,y), d(y,z))$. For a non-Archimedean metric, a sequence is Cauchy iff $|x_n - x_{n+1}| \to 0$ as $n \to \infty$.
To prove that $F$ is complete under this metric, note that for a Cauchy sequence in $F$, the coefficients for a fixed exponent $t^n$ must be eventually constant; otherwise the difference between consecutive terms could never be smaller than $q^{-n}$. Call this constant $a_n$; then the Cauchy sequence converges to $\sum a_n t^n$.
Best Answer
Consider the ring of formal Laurent series $R((x))$ with the ordering where $x$ is a positive infinitesimal.
That is, a rational function is positive if and only if its Laurent series has a positive leading coefficient.
What ($\omega$-indexed) sequences converge to zero?
Well, for some $n$, we must have $s_m < x^2$ for all $m > n$. In particular, this means the leading term cannot be of the form $a x^j$ with $j < 2$, because such a thing would be greater than $x^2$. So for all $m > n$, the coefficient on $x_j$ is $0$ for all $j < 0$.
A similar argument can be used in each degree; so we have a simple characterization of sequences that converge to zero: they are the sequences bounded on the left for which the sequence of coefficients on each $x^i$ is eventually always zero. (Note that it's not enough for the coefficients to simply converge to zero!)
The bounded criterion rules out things like the sequence
$$ 1, x^{-1}, x^{-2}, x^{-3}, \cdots $$
which diverges to $+\infty$.
Correspondingly, there is a simple condition for Cauchy sequences: they are precisely the sequences which are bounded on the left and for which each the sequence of coefficients on each $x^i$ is eventually constant.
Therefore, $R((x))$ is Cauchy complete (in the sense that every Cauchy sequence converges), and it is non-Archimedean because it has positive infinite numbers, such as $x^{-1}$.