= Solution
Define the <momentum density> $q_i=(\bar\rho+\bar P)v_{\perp i}$, lowering its spatial label with $\delta_{ij}$, so the mixed <stress-energy tensor> obeys $T^i{}_0=q_i$ and $T^0{}_i=-q_i$. All products of $J_{ij}$ or its <derivatives> with $q_i$ are second order and must be dropped in <linear cosmological perturbation theory>.
For $\nu=0$, the three terms in <stress-energy conservation> are
$$
\partial_\mu T^\mu{}_0=\bar\rho'+\partial_iq_i,\qquad
\Gamma^\mu_{\mu\sigma}T^\sigma{}_0=4\mathcal H\bar\rho,\qquad
\Gamma^\sigma_{\mu0}T^\mu{}_\sigma=\mathcal H\bar\rho-3\mathcal H\bar P.
$$
Here the trace $J'_{ii}$ vanishes, $\Gamma^i_{00}=\Gamma^0_{i0}=0$, and $\partial_iq_i=(\bar\rho+\bar P)\partial_iv_\perp^i=0$. Thus
$$
\boxed{\bar\rho'+3\mathcal H(\bar\rho+\bar P)=0.}
$$
For $\nu=i$, spatial <pressure> gradients vanish because this is a purely <vector cosmological perturbation>. The first two terms are
$$
\partial_\mu T^\mu{}_i=-q_i',\qquad
\Gamma^\mu_{\mu\sigma}T^\sigma{}_i=-4\mathcal Hq_i.
$$
In the remaining contraction, the two background-connection times <momentum density> terms cancel:
$$
\Gamma^0_{ji}T^j{}_0+\Gamma^j_{0i}T^0{}_j=\mathcal Hq_i-\mathcal Hq_i=0.
$$
The other terms are $\Gamma^0_{0i}\bar\rho=0$ and $-\bar P\Gamma^j_{ji}=0$. Consequently
$$
\boxed{[(\bar\rho+\bar P)v_{\perp i}]'+4\mathcal H(\bar\rho+\bar P)v_{\perp i}=0.}
$$
Expanding this <conservation law> and using the background <cosmological continuity equation> gives the equivalent <peculiar velocity> equation, when $\bar\rho+\bar P\ne0$,
$$
\boxed{v_{\perp i}'+\left(\mathcal H+\frac{\bar P'}{\bar\rho+\bar P}\right)v_{\perp i}=0.}
$$
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