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
$$
t_{\mathcal{R}} = t - \frac{\mathcal{R}} {c}
$$
The results are:
$$
\vec{S}\left(\vec{r},t\right)\cdot\hat{n}\left(\vec{r}\right) = 
  -
  \frac {
  \left( \vec{Y}\left(t_\mathcal{R}\right) \times \hat{r} \right)
  \cdot 
  \left(\vec{Z}\left(t_\mathcal{R}\right) \times \hat{r} \right) 
  } {16 \pi^2 \epsilon_0\mathcal{R}^3 c^2}
  - 
  \frac {
  \left( \vec{Z}\left(t_\mathcal{R}\right)  \times \hat{r} \right)
  \cdot 
  \left(\vec{Z}\left(t_\mathcal{R}\right) \times \hat{r} \right)
  } {16 \pi^2 \epsilon_0\mathcal{R}^2 c^3}
\\
P\left(\mathcal{R},t\right) =
  \frac {1} {6 \pi \epsilon_0 c^2}
  \left(
    \frac 
       {\vec{Y}\left(t_{\mathcal{R}}\right)\cdot\vec{Z}\left(t_{\mathcal{R}}\right)}
       {\mathcal{R}}
    +
    \frac 
       {\left|\vec{Z}\left(t_{\mathcal{R}}\right)\right|^2} 
       {c}
  \right)
\\
\mathcal{P}_{\infty}\left(t\right) =
  \frac 
    {\left|\vec{Z}\left(t\right)\right|^2} 
    {6 \pi \epsilon_0 c^3}
,\qquad
\mathcal{P}_{\delta}\left(\mathcal{R},t\right) =
    \frac 
       {\vec{Y}\left(t\right)\cdot\vec{Z}\left(t\right)}
       {6 \pi \epsilon_0 \mathcal{R} c^2}
$$

## 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.

For the system defined in the question, we designate the center of $V$ (and $V_s$) as $\vec{r}_0$. For all $\vec{r}\in\partial V$ and all $\vec{r}_s\in V_s$, we also define:
$$
\hat{n}\left(\vec{r}\right) = \frac {\vec{r}-\vec{r}_0} {\left|\vec{r}-\vec{r}_0\right|}
,\qquad
k\left(\vec{r}, \vec{r}_s\right) = \frac {R} {\mathcal{R}} =\frac {\left|\vec{r} - \vec{r}_s\right|} {\left|\vec{r} - \vec{r}_0\right|}, 
\qquad 
h\left(\vec{r}, \vec{r}_s\right) = 
  \frac {\left(\vec{r} - \vec{r}_s\right)} {\left|\vec{r} - \vec{r}_s\right|} 
  \cdot
  \hat{n}\left(\vec{r}\right)
$$
When $\mathcal{R} \gg \mathcal{R}_s$, it can be seen that $k\left(\vec{r}, \vec{r}_s\right)\approx 1$ and $h\left(\vec{r}, \vec{r}_s\right)\approx 1$ irrespective of $\vec{r}_s$. Under this 'far field' approximation, therefore, $\vec{R}\approx\vec{r}-\vec{r}_0$, 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 $\vec{S}\left(\vec{r},t\right)\cdot\hat{n}\left(\vec{r}\right)$

Now 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. The third term vanishes, and for a sphere $V$ of radius $\mathcal{R}$ we get:
$$
\oint_{\partial V} \sin^2\vartheta\left(\vec{r}\right) \space ds(\vec{r}) = \frac {8 \pi \mathcal{R}^2} {3}
$$

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:
$$
\int_{t1}^{t2} \vec{Y}(t)\cdot\vec{Z}(t) dt = \int_{t1}^{t2} \vec{Y}(t)\cdot\frac{d\vec{Y}}{dt}(t) dt = \frac12 \int_{t1}^{t2} \frac{d}{dt}|\vec{Y}(t)|^2 dt = \frac12 \left[|\vec{Y}(t)|^2\right]_{t1}^{t2}
$$
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 $\sin\vartheta\left(\vec{r},t_{\mathcal{R}}\right) = 0$, that is, wherever the unit normal is parallel to $\vec{Z}\left(t_{\mathcal{R}}\right)$