For $\mathcal{R} \gg \mathcal{R}_s$, if we define:
$$
\vec{Y}\left(t\right)=
\iiint_{V_s}
\vec{J}\left(\vec{r},t\right)
\space dV\left(\vec{r}\right)
,\qquad
\vec{Z}\left(t\right)=
\iiint_{V_s}
\frac{\partial \vec{J}\left(\vec{r},t\right)} {\partial t}
\space dV\left(\vec{r}\right)
= \frac {d \vec{Y}\left(t\right)} {dt}
$$
and also
$$
t_{\mathcal{R}} = t - \frac{\mathcal{R}} {c}
$$
Assuming that the origin is at the center of $V$, so that $\hat{n}\left(\vec{r}\right) = \hat{r}$, the results are:
$$
\vec{S}\left(\vec{r},t\right)\cdot\hat{n}\left(\vec{r}\right) =
-
\frac {
\left( \vec{Y}\left(t_\mathcal{R}\right) \times \hat{n}\left(\vec{r}\right) \right)
\cdot
\left(\vec{Z}\left(t_\mathcal{R}\right) \times \hat{n}\left(\vec{r}\right) \right)
} {16 \pi^2 \epsilon_0\mathcal{R}^3 c^2}
-
\frac {
\left| \vec{Z}\left(t_\mathcal{R}\right) \times \hat{n}\left(\vec{r}\right) \right|^2
} {16 \pi^2 \epsilon_0\mathcal{R}^2 c^3}
\\
P\left(\mathcal{R},t\right) =
\frac {1} {6 \pi \epsilon_0 c^2}
\left(
\frac
{\vec{Y}\left(t_{\mathcal{R}}\right)\cdot\vec{Z}\left(t_{\mathcal{R}}\right)}
{\mathcal{R}}
+
\frac
{\left|\vec{Z}\left(t_{\mathcal{R}}\right)\right|^2}
{c}
\right)
\\
\mathcal{P}_{\infty}\left(t\right) =
\frac
{\left|\vec{Z}\left(t\right)\right|^2}
{6 \pi \epsilon_0 c^3}
,\qquad
\mathcal{P}_{\delta}\left(\mathcal{R},t\right) =
\frac
{\vec{Y}\left(t\right)\cdot\vec{Z}\left(t\right)}
{6 \pi \epsilon_0 \mathcal{R} c^2}
$$
Derivation
Given the following definitions:
$$
\vec{R} = \vec{r} - \vec{r}_s, \qquad R = \left | \vec{R} \right |, \qquad t_r = t - \frac {R} {c}
$$
From Jefimenko's Equations, the value of $\vec{E}$ and $\vec{B}$ for any $\vec{r}$ and any $t$ is as follows:
$$
\vec{E}(\vec{r},t) =
\frac {1} {4 \pi \epsilon_0}
\iiint_{V_s} {\left(
\frac {\rho (\vec{r}_s, t_r)} {R^3} \vec{R} +
\frac {1} {R^2 c} \frac {\partial \rho (\vec{r}_s, t_r) } {\partial t} \vec{R} -
\frac {1} {R c^2} \frac {\partial \vec{J} (\vec{r}_s, t_r) } {\partial t} \right)}
\space dV\left(\vec{r}_s\right)
$$
$$
\vec{B}(\vec{r},t) =
\frac {\mu_0} {4 \pi}
\iiint_{V_s} {\left(
\frac {\vec{J} (\vec{r}_s, t_r)} {R^3} \times \vec{R} +
\frac {1} {R^2 c} \frac {\partial \vec{J} (\vec{r}_s, t_r) } {\partial t} \times \vec{R} \right)}
\space dV\left(\vec{r}_s\right)
$$
Using this, we can write down the expression for $\vec{S}\left(\vec{r},t\right)$, which has six terms, each of which is a product of two volume integrals. We are, however, interested in $\vec{S}\left(\vec{r},t\right) \cdot \hat{n}\left(\vec{r}\right)$, where, if we designate the center of $V$ (and $V_s$) as $\vec{r}_0$:
$$
\hat{n}\left(\vec{r}\right) = \frac {\vec{r}-\vec{r}_0} {\left|\vec{r}-\vec{r}_0\right|}
$$
Now if we choose a frame of reference where $\vec{r}_0$ is the origin, we find that $\hat{n}\left(\vec{r}\right) = \hat{r}$.
When $\mathcal{R} \gg \mathcal{R}_s$, we can consider $\dfrac{\left|\vec{r}_s\right|}{\left|\vec{r}_s\right|} \approx 0$. Thus, irrespective of $\vec{r}_s$ under this 'far field' approximation, $\vec{R}\approx\vec{r}-\vec{r}_s$, and is effectively independent of $\vec{r}_s$, so it can be taken outside of the volume integrals.
Then, because $\left(\vec{a}\times\left(\vec{b}\times\vec{a}\right)\right)\cdot\vec{a}=0$ four of the six terms in $\vec{S}\left(\vec{r},t\right)\cdot\hat{n}\left(\vec{r}\right)$ turn out to be zero.
Next, we substitute $\vec{Y}$ and $\vec{Z}$ into the expression for $\vec{S}\left(\vec{r},t\right)\cdot\hat{n}\left(\vec{r}\right)$ and get the expression listed above.
To help us with the surface integral, we define a spherical coordinate system $(\mathscr{r},\theta,\phi),\space 0\le \mathscr{r}\lt\infty,\space 0\le\theta\le\pi,\space 0\le\phi\le 2\pi$ with its center at $\vec{r}_0$, and orient it such that:
$\vec{Z}\left(t\right)$ is $( |\vec{Z}\left(t\right)|, 0, 0)$ and
$\vec{Y}\left(t\right)$ is $( |\vec{Y}\left(t\right)|, \gamma\left(t), 0\right)$.
Note that the orientation of this coordinate system changes with time. In this coordinate system, therefore, $\hat{n}\left(\vec{r}\right)$ at any instant $t$ has to be represented in terms of the retarded time as
$\left(1,\vartheta\left(\vec{r},t_r\right), \varphi\left(\vec{r},t_r\right)\right)$
Working out $\vec{S}\left(\vec{r},t\right)\cdot\hat{n}\left(\vec{r}\right)$ in this coordinate system, we get:
$$
\vec{S}\left(\vec{r},t\right)\cdot\hat{n}\left(\vec{r}\right) =
\frac {\left|\vec{Z}\left(t_{\mathcal{R}}\right)\right|^2} {16 \pi^2 \epsilon_0 {\mathcal{R}}^2 c^3}
\sin^2\vartheta\left(\vec{r},t_{\mathcal{R}}\right)
+ \frac {\vec{Y}\left(t_{\mathcal{R}}\right)\cdot\vec{Z}\left(t_{\mathcal{R}}\right)} {16 \pi^2 \epsilon_0 {\mathcal{R}}^3 c^2}
\sin^2\vartheta\left(\vec{r},t_{\mathcal{R}}\right) \\
- \frac {\left|\vec{Y}\left(t_{\mathcal{R}}\right)\times\vec{Z}\left(t_{\mathcal{R}}\right)\right|} {32 \pi^2 \epsilon_0 {\mathcal{R}}^3 c^2}
\sin2\vartheta\left(\vec{r},t_{\mathcal{R}}\right)
\cos\varphi\left(\vec{r},t_{\mathcal{R}}\right)
$$
Finally, we compute the surface integral. For a sphere $V$ of radius $\mathcal{R}$ we get:
$$
\oint_{\partial V} \sin^2\vartheta\left(\vec{r}\right) \space ds(\vec{r}) = \frac {8 \pi \mathcal{R}^2} {3},\qquad
\oint_{\partial V} \sin 2\vartheta\left(\vec{r}\right) \space\cos\varphi\left(\vec{r}\right) \space ds(\vec{r})= 0
$$
Substituting, we get our result.
Observations
For the record,because of the way we've defined $V_s$ and because of conservation of charge, $Q_s$ is the (constant) total charge inside $V_s$,
$$
\iiint_{V_s}
\rho\left(\vec{r},t\right)
\space dV\left(\vec{r}\right) = Q_s
,\qquad
\iiint_{V_s}
\frac{\partial \rho\left(\vec{r},t\right)} {\partial t}
\space dV\left(\vec{r}\right)
= \frac {d Q_s} {dt}
= 0
$$
... but we don't need these terms in the derivation.
Also, $\int_{t_1}^{t_2} \mathcal{P}_{\delta}\left(\mathcal{R},t\right) \space dt = 0$ because:
$$
\int_{t_1}^{t_2} \vec{Y}(t)\cdot\vec{Z}(t) dt = \int_{t_1}^{t_2} \vec{Y}(t)\cdot\frac{d\vec{Y}}{dt}(t) dt = \frac12 \int_{t_1}^{t_2} \frac{d}{dt}|\vec{Y}(t)|^2 dt = \frac{1}{2} \left[|\vec{Y}(t)|^2\right]_{t_1}^{t_2}
$$
and $$\forall\vec{r}\in V: \vec{J}\left(\vec{r},t_1\right) = \vec{J}\left(\vec{r},t_2\right) \implies \vec{Y}\left(t_1\right) = \vec{Y}\left(t_2\right)$$
The value of $\vec{S}\left(\vec{r},t\right)\cdot\hat{n}\left(\vec{r}\right)$ is zero whenever $\vec{Z}\left(t_{\mathcal{R}}\right) \times \hat{n}\left(\vec{r}\right) = 0$, that is, wherever the unit normal is parallel to $\vec{Z}\left(t_{\mathcal{R}}\right)$
