The $k$-Hodge conjecture is false for any $k$ containing an irrational real $\alpha$.

Indeed we may clearly assume $\alpha$ is negative. Consider the elliptic curve $E = \mathbb C / \langle 1, \sqrt{\alpha} \rangle$. Let $e_1$ and $e_1$ be a basis of $H^1(E,\mathbb Z)$ such that the map to $H^{1,0}(E)$ sends $e_1$ to $1$ and $e_2$ to $\sqrt{\alpha}$. Then in $H^1(E,k) \otimes H^1(E,k) \subset H^2(E \times E, k)$, consider the class $$\alpha( e_1 \otimes e_1) - (e_2 \otimes e_2)$$

In the natural projection to $H^{2,0}$, it is sent to $\alpha(1) - (\sqrt{\alpha})^2=0$. In the natural projection to $H^{2,0}$, which is the complex conjugate of that, it is sent to $\alpha (1) - (-\sqrt{\alpha})^2=0$. So it lies in $H^{1,1}(E\times E) $.

But because $E$ is not CM (because $\alpha$ is irrational $\sqrt{\alpha}$ is not a quadratic irrational), the only algebraic cycle that contributes to $H^1(E) \otimes H^1(E)$ is the diagonal, which has class $e_1 \otimes e_2 - e_2 \otimes e_1$, and this class is not a multiple of it.

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On the other hand, assuming the $\mathbb Q$-Hodge conjecture, we obtain the $k$-Hodge conjecture for any imaginary quadratic field $k$. Indeed, for $k= \mathbb Q(\sqrt{-D})$, $H^{2q}(X,\mathbb Q(\sqrt{-D}))= H^{2q}(X,\mathbb Q) + \sqrt{-D}H^{2q}(X,\mathbb Q)$.   Because complex conjugation exchanges $H^{q,p}$ and $H^{p,q}$, it fixes $H^{q,q}$, so if a class in $H^{2q}(X,\mathbb Q(\sqrt{-D}))$ lands in $H^{q,q}$, both its real and imaginary parts do as well, so under the Hodge conjecture they are $\mathbb Q$-linear combinations of algebraic cycles, and the original is clearly a $\mathbb Q(\sqrt{-D})$-linear combination of algebraic cycles.

I'm not sure about number fields of degree at least $3$ that contain no real irrationals.