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} $.
Best Answer
Since the phrasing of the question itself is a bit muddled, let me instead address the version of the question you stated in a comment:
The answer is absolutely yes (assuming that by "axioms" you mean an axiomatic description of a structure in an ambient set theory, rather than some kind of self-contained first-order axiomatization). In particular, the real numbers can be completely axiomatized as a complete ordered field: that is, a field $\mathbb{R}$ together with a total ordering compatible with the field structure, such that any nonempty set $S\subset\mathbb{R}$ with an upper bound has a least upper bound.
Let me sketch a proof from these axioms that $\mathbb{R}$ is uncountable. Suppose $\mathbb{R}$ were countable, and $i:\mathbb{R}\to\mathbb{N}$ were an injection. We have elements $0,1\in\mathbb{R}$ with $0<1$. Moreover, the ordered field axioms imply that if $x,y\in\mathbb{R}$ satisfy $x<y$, then there exists an element $z\in\mathbb{R}$ such that $x<z<y$ (namely, $z=(x+y)/2)$. Now define an increasing sequence $(x_n)$ and a decreasing sequence $(y_n)$ such that $x_n<y_n$ for all $n$ by induction. Start with $x_0=0$ and $y_0=1$. Given $x_n$ and $y_n$ with $x_n<y_n$, consider the set of all $z\in\mathbb{R}$ such that $x_n<z<y_n$. This set is nonempty, and so it has a unique element $z$ which minimizes the value $i(z)$. Let $x_{n+1}$ be that value $z$. Similarly, there is a unique $w$ such that $x_{n+1}<w<y_n$ which minimizes the value $i(w)$ among all such $w$. Let $y_{n+1}=w$.
Having constructed these sequences, let $S=\{x_n\}_{n\in\mathbb{N}}$. This is a nonempty subset of $\mathbb{R}$, and it is bounded above (namely, by $y_n$ for any $n$). So there is some $r\in \mathbb{R}$ that is the least upper bound of the set $S$. Note that the numbers $y_n$ are all distinct, so the numbers $i(y_n)$ form an infinite subset of the natural numbers. In particular, there must be some $n$ such that $i(y_{n+1})>i(r)$. But I claim this is impossible. Indeed, $y_{n+1}$ was defined to be the element $w\in\mathbb{R}$ such that $x_{n+1}<w<y_n$ which minimized $i(w)$. But $r$ also satisfies $x_{n+1}<r<y_n$ ($r\geq x_{n+2}>x_{n+1}$ since $r$ is an upper bound of $S$, and $r\leq y_{n+1}<y_n$ since $r$ is the least upper bound of $S$ and $y_{n+1}$ is also an upper bound of $S$). Since $i(r)<i(y_{n+1})$, this contradicts our definition of $y_{n+1}$. This contradiction means that our injection $i$ cannot exist, i.e. $\mathbb{R}$ is uncountable.
For an alternate proof, you can define decimal expansions of real numbers from these axioms (though it takes a bit of work), and then do the usual diagonal argument.