= Solution
Only the $1/r$ fields transport a nonzero outgoing <power> to infinity. Retain those terms from the <Hertzian dipole> fields:
$$
E_\theta^{\rm rad}=\frac{\sin\theta}{4\pi\epsilon_0c^2r}[\ddot p],\qquad
B_\phi^{\rm rad}=\frac{\mu_0\sin\theta}{4\pi cr}[\ddot p]
=\frac{E_\theta^{\rm rad}}c.
$$
The brackets mean evaluation at the <retarded time>,
$$
[\ddot p]\equiv\ddot p(t-r/c),
$$
for a point source at the origin in <vacuum>. They are not a time average. The radial <Poynting vector> in the <radiation zone> is
$$
S_r=\frac1{\mu_0}E_\theta^{\rm rad}B_\phi^{\rm rad}
=\frac{\mu_0}{16\pi^2cr^2}[\ddot p]^2\sin^2\theta.
$$
Integrating over a <sphere> uses
$$
\int\sin^2\theta\,d\Omega=2\pi\int_0^\pi\sin^3\theta\,d\theta=\frac{8\pi}3.
$$
Therefore
$$
\boxed{P(t,r)=\int r^2S_r\,d\Omega=\frac{\mu_0}{6\pi c}[\ddot p]^2.}
$$
Near-field terms either give inverse powers of $r$ after the surface integral or describe reversible exchange with the near field. The emitted <power> labelled by source time $t_{\rm src}$ is $\mu_0\ddot p(t_{\rm src})^2/(6\pi c)$; the enclosing surface receives it one light-travel time later. For a changing dipole orientation, the general form is $P=\mu_0|\ddot{\mathbf p}|^2/(6\pi c)$.
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