There is a worked-out proof in page 10 and following of <A HREF="https://hal.archives-ouvertes.fr/hal-01616979/document">Hérau's lecture notes.</A> The detailed steps are for $n=1$, $V=0$, but I assume once that is understood, the more general case would follow smoothly.
    
As a short-hand notation we write $||\partial_x h||^2=\int (\partial h/\partial x)^2\,d\mu$ and $\langle\partial_x h,\partial_v h\rangle=\int (\partial h/\partial x)(\partial h/\partial v)\,d\mu$. We will make use of the identities
$$\langle f,\partial_v g\rangle=\langle(-\partial_v+v)f,g\rangle,$$
$$\langle v\partial_x f,f\rangle=0$$
$$[\partial_v v,\partial_x]=\partial_x,$$
and the Cauchy-Schwartz inequality
$$2|c\langle\partial_v f,\partial_x f\rangle|\leq c^2||\partial_v f||^2+||\partial_x f||^2.$$
The Fokker-Planck equation for $n=1$, $V=0$ reads
$$\partial_t h+v\partial_x h+(-\partial_v+v)\partial_v h=0.$$
This implies the derivatives
$$-\frac{1}{2}\frac{d}{dt}||h||^2=||\partial_v h||^2,$$
$$-\frac{1}{2}\frac{d}{dt}||\partial_x h||^2=||\partial_v\partial_x h||^2,$$
$$-\frac{1}{2}\frac{d}{dt}||\partial_v h||^2=\langle\partial_x h,\partial_v h\rangle+||(-\partial_v +v)\partial_v h||^2=\langle\partial_x h,\partial_v h\rangle+||\partial_v^2 h||^2+||\partial_v h||^2.$$

So the derivative $dI/dt$ in the OP reduces to
$$\frac{dI}{dt}=-2a||\partial_v\partial_x h||^2-2c\biggl(\langle\partial_x h,\partial_v h\rangle+||\partial_v^2 h||^2+||\partial_v h||^2\biggr)$$
$$\qquad\leq -2a||\partial_v\partial_x h||^2-2c||\partial_v^2 h||^2-(2c-c^2)||\partial_v h||^2+||\partial_x h||^2.$$
It remains to bound $||\partial_x h||^2$. 

In the lecture notes they do this by adding the mixed term $b\langle\partial_x h,\partial_v h\rangle$ to the left-hand-side of the inequality, which then gives a term $-b||\partial_x h||^2$ on the right-hand-side to dominate. Let me work that out, using the derivative
$$\frac{d}{dt}\langle\partial_x h,\partial_v h\rangle=-||\partial_x h||^2+2\langle(-\partial_v+v)\partial_v h,\partial_v\partial_x h\rangle-\langle\partial_x h,\partial_v h\rangle,$$
hence
$$\frac{d}{dt}J\equiv\frac{d}{dt}\biggl(||h||^2+a||\partial_x h||^2+b\langle\partial_x h,\partial_v h\rangle+c||\partial_v h||^2\biggr)=$$
$$\qquad=-2||\partial_v h||^2-2a||\partial_v\partial_x h||^2-b||\partial_x h||^2-2c||(-\partial_v +v)\partial_v h||^2+2b\langle(-\partial_v+v)\partial_v h,\partial_v\partial_x h\rangle-(b+2c)\langle\partial_x h,\partial_v h\rangle.$$
The first four terms on the right-hand-side have a fixed sign, the last two terms are bounded by the Cauchy-Schwartz inequality,
$$\frac{d}{dt}J\leq -(2a-1)||\partial_v\partial_x h||^2-(b-1/2)||\partial_x h||^2-(2c-b^2)||(-\partial_v +v)\partial_v h||^2-(1-c-b/2)||\partial_v h||^2.$$

For $b=1/2$ we thus find
$$\frac{d}{dt}J\leq-(2a-1)||\partial_v\partial_x h||^2-(2c-1/4)||\partial_v^2 h||^2-(1/2+c)||\partial_v h||^2$$
$$\qquad \leq-K\biggl(||\partial_v\partial_x h||^2+||\partial_v^2 h||^2+||\partial_v h||^2\biggr)$$
for some $K$ if $a>1/2$, $c>1/8$.