Yes and no.
Why "no"?
Technically direct current describes a stationary situation, when the current is constant in time. This is obviously not the case just after the switch is turned on, i.e., when the current grows from zero to its steady value. This is also not the case when the switch is turned off and the current decays. To describe the current growing from zero to steady value one would need to use a more refined model than just a resistor conencted to a battery - including wire capacitance (and possibly also its inductance) would do. (Note that any wire has capacitance and inductance, it is just that for short wires these are often neglected, but it is not the case for the electric transmission lines or the microwires on an electronic chip).
When "yes"?
If we consider a straight infinitely long wire, parallel to the $z$-axis and carrying current $I(t)$, we can write the current density as
$$
\mathbf{J}(\mathbf{r},t) = \delta(x)\delta(y)I(t)\mathbf{e}_z.
$$We can now substitute this current to the Maxwell equations (or better into the electromagnetic wave equations that follow from the Maxwell equations) and solve for the resulting field. Substituting the expressions above into the general solution for the vector potential we have:
$$
\mathbf{A}(\mathbf{r}, t)=
\int\frac{\delta\left(t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}-t\right)}{|\mathbf{r}-\mathbf{r}'|}
\frac{\mathbf{J}(\mathbf{r}', t')}{c}d\mathbf{r}'dt'
=
\frac{\mathbf{e}_z}{c}\int \frac{I\left(t+\frac{\sqrt{x^2+y^2+(z-z')^2}}{c}\right)}{\sqrt{x^2+y^2+(z-z')^2}}dz'
$$
If the current is time-independent, i.e., $I(t)$I_0$, the vector potential is idnependent on time and no electromagnetic waves generated. However, in case of time-dependent current we can expand the equation above itno Fourrier integral, i.e., into a spectrum of electromagnetic waves.
Remarks:
The solution above is simplistic. In real situations there are other factors to account for, such as:
- the wire is not necessarily straight, and not infinite - the other parts of the circuit are also emitting
- Whether the EM wave actually propagates and is detectable depends on its intensitya nd the environment (typical circuits emit waves that are too weak and quickly decay).
- One needs to account for the feedback - the change of the current in the circuit due to the rpesence of the wave, i.e., one really has to solve a self-consistent problem.