This is a countable family of first-order statements, so it holds for every real-closed field, since it holds over $\mathbb R$.
From a square matrix, we immediately derive that such a field must satisfy the property that the sum of two perfect squares is a perfect square. Indeed, the matrix:
$ \left(\begin{array}{cc} a & b \\ b & -a \end{array}\right)$
has characteristic polynomial $x^2-a^2-b^2$, so it is diagonalizable as long as $a^2+b^2$ is a pefect square.
Moreover, $-1$ is not a perfect square, or else the matrix:
$ \left(\begin{array}{cc} i & 1 \\ 1 & -i \end{array}\right)$
would be diagonalizable, thus zero, an obvious contradiction.
So the semigroup generated by the perfect squares consists of just the perfect squares, which are not all the elements of the field, so the field can be ordered. It is not immediately obvious to me whether the field must necessarily be real closed according to this order.