A field without a canonical square root of $-1$

$\overline{\mathbb{Q}_5}$. The square roots of $-1$ actually already lie in $\mathbb{Z}_5$ and can be computed using Hensel's lemma but there's no reason to prefer the one congruent to $2 \bmod 5$ over the one congruent to $3 \bmod 5$ or vice versa.

I am not sure this would pass the test as it is possible that most mathematicians would give you the one congruent to $2 \bmod 5$ just because it is the first square root of $-1$ in $\mathbb{Z}/5\mathbb{Z}$ you find as you start from $0$ and add $1$, so take a larger prime congruent to $1 \bmod 4$ instead of $5$, maybe, to force them to try something smarter (say a prime large enough that it is not feasible to compute $\left( \frac{p-1}{2} \right)! \bmod p$).

You can model the complex numbers by linear combinations of the $2\times 2$ unit matrix $\mathbb{I}$ and a real $2\times 2$ skew-symmetric matrix with square $-\mathbb I$, of which there are two, $\begin{pmatrix}0 & -1\\1 & 0\end{pmatrix}$ and $\begin{pmatrix}0 & 1\\-1 & 0\end{pmatrix}$. I see no obvious reason to prefer one over the other.

I don't feel that the question is really a coherent one. Let me try to illustrate that as follows.

Suppose I hand you the field $\overline{\mathbb{C}} = \{a - bi \ | \ a+bi \in \mathbb{C} \}$. It seems that in your sense of canonical, the canonical square root of $-1$ in this field is $-i$.

But wait: $\overline{\mathbb{C}}$ is not a(n even set-theoretically) different field from $\mathbb{C}$! It is just being presented differently: namely starting with the incarnation you usually have of $\mathbb{C}$ (which you have not specified in your question, but is rather implicit in its formulation that you have one) and compose it with the complex conjugation automorphism. The point here is that it doesn't make sense to distinguish between the fields $\mathbb{C}$ and $\overline{\mathbb{C}}$, only to distinguish between different identifications of these fields.

In summary, whenever your field has an automorphism which moves an element $x$ such that $x^2 = -1$, there is no canonical choice between $x$ and $-x$, at least not in any mathematically robust sense.

A field abstractly isomorphic to $\mathbb{C}$ but does not have a canonical choice of square root of $-1$ is $\mathbb{C}$ itself. There is nothing canonical about which square root of $-1$ to call $i$ and which square root to call $-i.$ By calling one of the square roots $i$ you have just set up notation, nothing more. The fact that there is a field automorphism of $\mathbb{C}$ which interchanges the two square roots confirms this, as it preserves all algebraic structure. Even more, it preserves the natural metric structure of $\mathbb{C}.$

The answer to one possible formalization of the question is yes. It is consistent that there is a definable field isomorphic to $\mathbb{C}$ with no definable imaginary unit. To construct such a thing, we need a pair of sets $\{X,Y\}$ that is definable, but such that neither $X$ nor $Y$ is definable. It is consistent for such a strange object to exist: see François's answer to the question Does every nonempty definable finite set have a definable member?)

Now let $\mathbb{C}$ be your favorite construction of the complex field. One of its imaginary units, $i$, is definable, or else we are done. Define the structure $F = (\mathbb{C} \times \{X,Y\}) / \mathord{\sim}$ where $(a+bi, X) \sim (a-bi, Y)$ for all $a,b \in \mathbb{R}$. Then $F$ is a definable structure isomorphic to $\mathbb{C}$ and its imaginary units are $\{(i,X),(-i,Y)\}$ and $\{(-i,X),(i,Y)\}$. Neither of these imaginary units can be definable, or else $X$ and $Y$ would be definable from $i$ and therefore definable.