Not too sure just what "canonical form" means in this context, but with respect to general matrix terminology, the phrase is usually used to refer to some sort of diagonal, or "near diagonal", representation, e.g., "Jordan canonical form", "rational canonical form", and so forth. In most of these "forms", the diagonal of a matrix plays a central role. Bearing these thoughts in mind, I offer the following words on the matter:
We have a unitary $U$:
$UU^\dagger = U^\dagger U = I, \tag 1$
which is also self-adjoint:
$U = U^\dagger; \tag 2$
(2) implies $U$ may be diagonalized by some unitary $W$:
$W^\dagger U W = \Lambda = \text{diag}(\mu_1, \mu_2, \ldots, \mu_n), \tag 3$
with
$W^\dagger W = WW^\dagger = I; \tag 4$
the $\mu_i$ are the eigenvalues of $U$; from (2), they are all real; therefore
$\Lambda^\dagger = \Lambda; \tag 5$
from (3),
$\Lambda^\dagger = W^\dagger U^\dagger W; \tag 6$
and thus,
$\Lambda^2 = \Lambda^\dagger \Lambda = W^\dagger U^\dagger W W^\dagger U W; \tag 7$
since we have (2) and (4),
$\Lambda^2 = W^\dagger U^\dagger IU W = W^\dagger U^\dagger U W =W^\dagger I W =W^\dagger W = I; \tag 8$
also,
$\Lambda^2 = \text{diag}(\mu_1^2, \mu_2^2, \ldots, \mu_n^2); \tag 9$
thus,
$\mu_i^2 = 1 \Longrightarrow \mu_i = \pm 1, \; 1 \le i \le n; \tag{10}$
thus, $\Lambda$ must be a matrix of the form
$\Lambda = \text{diag}(\pm 1, \pm 1, \ldots, \pm 1); \tag{11}$
and hence $U$ is of the general form
$U = W\Lambda W^\dagger = W \text{diag}(\pm 1, \pm 1, \ldots, \pm 1)W^\dagger, \tag{12}$
for some choice of $\pm 1$ on the diagonal of $\Lambda$, and for some appropriate choice of unitary $W$.
It should be clear that any matrix given by (12) satisfies (1) and (2).
This representation is as "canonical" as I know how to make it!