Preamble
I believe that the OP is seeking a characterization of $ \mathbb{Q} $ using only the first-order language of fields, $ \mathcal{L}_{\text{Field}} $. Restricting ourselves to this language, we can try to uncover new axioms, in addition to the usual field axioms (i.e., those that relate to the associativity and commutativity of addition and multiplication, the distributivity of multiplication over addition, the behavior of the zero and identity elements, and the existence of a multiplicative inverse for each non-zero element), that describe $ \mathbb{Q} $ uniquely.
Any attempt to describe the smallest field satisfying a given property must prescribe a method of comparing one field with another (namely using field homomorphisms, which are injective if not trivial), but such a method clearly cannot be formalized using $ \mathcal{L}_{\text{Field}} $.
1. There Exists No First-Order Characterization of $ \mathbb{Q} $
The answer is ‘no’, if one is seeking a first-order characterization of $ \mathbb{Q} $. This follows from the Upward Löwenheim-Skolem Theorem, which is a classical tool in logic and model theory.
Observe that $ \mathbb{Q} $ is an infinite $ \mathcal{L}_{\text{Field}} $-structure of cardinality $ \aleph_{0} $. The Upward Löwenheim-Skolem Theorem then says that there exists an $ \mathcal{L}_{\text{Field}} $-structure (i.e., a field) $ \mathbb{F} $ of cardinality $ \aleph_{1} $ that is an elementary extension of $ \mathbb{Q} $. By definition, this means that $ \mathbb{Q} $ and $ \mathbb{F} $ satisfy the same set of $ \mathcal{L}_{\text{Field}} $-sentences, so we cannot use first-order logic to distinguish $ \mathbb{Q} $ and $ \mathbb{F} $. In other words, as far as first-order logic can tell, these two fields are identical (an analogy may be found in point-set topology, where two distinct points of a non-$ T_{0} $ topological space can be topologically indistinguishable). However, $ \mathbb{Q} $ and $ \mathbb{F} $ have different cardinalities, so they are not isomorphic. This phenomenon is ultimately due to the fact that the notion of cardinality cannot be formalized using $ \mathcal{L}_{\text{Field}} $. Therefore, any difference between the two fields can only be seen externally, outside of first-order logic.
2. Finding a Second-Order Characterization of $ \mathbb{Q} $
This part is inspired by lhf's answer below, which I believe deserves more credit. We start by formalizing the notion of proper subfield using second-order logic.
Let $ P $ be a variable for unary predicates. Consider the following six formulas:
\begin{align}
\Phi^{P}_{1} &\stackrel{\text{def}}{\equiv} (\exists x) \neg P(x); \\
\Phi^{P}_{2} &\stackrel{\text{def}}{\equiv} P(0); \\
\Phi^{P}_{3} &\stackrel{\text{def}}{\equiv} P(1); \\
\Phi^{P}_{4} &\stackrel{\text{def}}{\equiv} (\forall x)(\forall y)((P(x) \land P(y)) \rightarrow P(x + y)); \\
\Phi^{P}_{5} &\stackrel{\text{def}}{\equiv} (\forall x)(\forall y)((P(x) \land P(y)) \rightarrow P(x \cdot y)); \\
\Phi^{P}_{6} &\stackrel{\text{def}}{\equiv} (\forall x)((P(x) \land \neg (x = 0)) \rightarrow (\exists y)(P(y) \land (x \cdot y = 1))).
\end{align}
What $ \Phi^{P}_{1},\ldots,\Phi^{P}_{6} $ are saying is that the set of all elements of the domain of discourse that satisfy the predicate $ P $ forms a proper subfield of the domain. The domain itself will be a field if we impose upon it the first-order field axioms. Hence,
$$
\{ \text{First-order field axioms} \} \cup \{ \text{First-order axioms defining characteristic $ 0 $} \} \cup \{ \neg (\exists P)(\Phi^{P}_{1} ~ \land ~ \Phi^{P}_{2} ~ \land ~ \Phi^{P}_{3} ~ \land ~ \Phi^{P}_{4} ~ \land ~ \Phi^{P}_{5} ~ \land ~ \Phi^{P}_{6}) \}
$$
is a set of first- and second-order axioms that characterizes $ \mathbb{Q} $ uniquely because of the following two reasons:
Up to isomorphism, $ \mathbb{Q} $ is the only field with characteristic $ 0 $ that contains no proper subfield.
If $ \mathbb{F} \ncong \mathbb{Q} $ is a field with characteristic $ 0 $, then $ \mathbb{F} $ does not model this set of axioms. Otherwise, interpreting “$ P(x) $” as “$ x \in \mathbb{Q}_{\mathbb{F}} $” yields a contradiction, where $ \mathbb{Q}_{\mathbb{F}} $ is the copy of $ \mathbb{Q} $ sitting inside $ \mathbb{F} $.
Any ordered field $F$ has characteristic $0$, so it contains a copy of $\mathbb{Z}$; by the universal property of the quotient field, the ring monomorphism $\mathbb{Z}\to F$ lifts to a monomorphism $\mathbb{Q}\to F$. We can identify $\mathbb{Q}$ with its image, so it's not restrictive to assume that $\mathbb{Q}\subseteq F$.
It's not really difficult: if $m/n\in\mathbb{Q}$, then we send it to
$$
\frac{f(m)}{f(n)}\in F
$$
where $f\colon \mathbb{Z}\to F$ is the (unique) monomorphism. Is this a field homomorphism? Just a check.
Now we come to the order. First of all, positive integers are positive in $(F,\prec)$: if $n>0$, then
$$
n=\underbrace{1+1+\dots+1}_{\text{$n$ times}}
$$
and therefore $0\prec n$. Conversely, if $n<0$, then
$$
n=-(\,\underbrace{\,1+1\dots+1}_{\text{$-n$ times}}\,)
$$
and so $n\prec0$.
Any element of $\mathbb{Q}$ can be represented as $m/n$ with $n>0$, because $a/b=(-a)/(-b)$, where $a,b\in F$, $b\ne0$. So, let $0\prec m/n$ in the ordering of $F$, with $n>0$. Then, by the properties of ordered fields,
$$
0\prec n\cdot\frac{m}{n}=m
$$
and therefore $m>0$. So a rational which is positive in $(F,\prec)$ is also positive in the usual order. A rational which is negative in $(F,\prec)$ is the opposite of a positive rational (in both orders).
Best Answer
Well, this depends on exactly what you mean by "the minimal ordered field". Let's say that we axiomatically define $\mathbb{Q}$ as an ordered field which contains no proper subfields.
Now, given such an ordered field $\mathbb{Q}$, let $K\subseteq\mathbb{Q}$ be the set of elements that can be written as a quotient of two integers (where we identify integers with elements of $\mathbb{Q}$ by taking sums of copies of $1$ or $-1$). Then it is easy to verify that $K$ is a subfield of $\mathbb{Q}$, and thus $K=\mathbb{Q}$. So for any $x\in\mathbb{Q}$, there are integers $a$ and $b$ such that $x=\frac{a}{b}$. Replacing $a$ and $b$ with their negatives if necessary, we may assume $b>0$. Since $b$ is an integer, this implies $b\geq 1$ (since $b$ is a sum of copies of $1$ and $1>0$). Thus $x=\frac{a}{b}\leq \frac{a}{1}=a$. Since $x\in\mathbb{Q}$ was arbitrary and $a$ is an integer, this proves $\mathbb{Q}$ is Archimedean.