For the second candidate, I assume that the Wigner distribution $ W(q,p) $ is smooth and continuous. This assumption was implicitly assumed with the first candidate entropy.
Given the property $ \lvert W(q,p) \rvert \le 2/h = 1/\pi \hbar$, that $ W(q,p) $ is bounded, and the properties you mentioned of how $ W(q,p) $ relates to $ \langle q \lvert \hat \rho \rvert q \rangle $ and $ \langle p \lvert \hat \rho \rvert p \rangle $, we should have that
$$ W(q,p) \le \int W(q,p) dp = \langle q \lvert \hat \rho \rvert q \rangle , $$ $$ W(q,p) \le \int W(q,p) dq = \langle p \lvert \hat \rho \rvert p \rangle . $$
With the property $ W^*(q,p) = W(q,p) $ and the property I mentioned of how $ W(q,p) $ relates to the expectation value of an operator $ \hat G $, we have the property
$$ 2 \pi \hbar \int \int W(q,p)^2 dq dp = \mathrm {Tr} ( \hat \rho^2 ) \le 1 . $$
Noting that $ \int \langle q \lvert \hat \rho \rvert q \rangle dq = \int \langle p \lvert \hat \rho \rvert p \rangle dp = \mathrm {Tr} ( \hat \rho ) = 1 $, then
$$ \int \int \langle q \lvert \hat \rho \rvert q \rangle \langle p \lvert \hat \rho \rvert p \rangle dq dp = \mathrm {Tr} ( \hat \rho )^2 = 1 . $$
Given that this integrand is in phase space, it stands to reason that we should have a similar inequality to the Cauchy-Schwarz,
$$ 2 \pi \hbar W (q,p)^2 \le \langle q \lvert \hat \rho \rvert q \rangle \langle p \lvert \hat \rho \rvert p \rangle . $$
Referring to the conjecture stated with the first candidate entropy and again using the convex function
$$ f(x) = - x \ln x , $$
we may then have
$$ S_{W^2} \le S_q + S_p , $$
where $ S_q $ and $ S_p $ are defined in the same manner as with the first candidate entropy, and
$$ S_{W^2} = \int \int f ( 2 \pi \hbar W(q,p)^2 ) dq dp = - \ln (2 \pi \hbar) \mathrm {Tr} (\hat \rho^2 ) - 2 \pi \hbar \int \int f ( W(q,p)^2 ) dq dp . $$
If this can be called an entropy, then Cauchy-Schwarz inequality needs to be proven. But as said, this is an attempt at finding and entropy which explicitly uses $ W(q,p) $.
Again reading the question, there may be a third candidate, also explicitly relating to $ W(q,p) $. Given what was stated about separability when $ \langle q \lvert \hat \rho \rvert q \rangle $ is Gaussian, then
$$ W(q,p) \le \lvert W(q,p) \rvert \le \langle q \lvert \hat \rho \rvert q \rangle \langle p \lvert \hat \rho \rvert p \rangle $$
and it may also be the case that
$$ S_{ \lvert W \rvert } \le S_q + S_p , $$
where
$$ S_{ \lvert W \rvert } = \int \int f ( \lvert W(q,p) \rvert ) dq dp . $$
However, since $ W(q,p) $ isn't necessarily positive definite, we could have that $ \int \lvert W(q,p) \rvert dq dp > 1 $, which means it may be possible to have $ S_{\lvert W \rvert} < 0 $. Perhaps, then, we should also consider $ - \int \int W(q,p) \ln \lvert W(q,p) \rvert dq dp $.