As of now, the best I got is:

$$
1 - e^{H_{1/2}(p)} \leq G(p) \leq \sqrt{1-e^{2 H_{1/2}(p)}},
$$
where $e^{H_{1/2}(p)} = 2 \ln \int_0^1p^{1/2}(x)dx$.

Both inequalities are saturated for both $G(p)=0$ and $G(p)=1$.

The lower bound is better than the previous one ($1 - e^{H_{0}(p)}$), as Rényi entropy $H_q(p)$ is non-decreasing in $q$.

The higher bound is different from $\sqrt{\tfrac{1}{3}e^{-H_2(p)}-1}$ (there is no inequality between them).

And for practical purposes their quadratic mean seems to be a good approximation,
$$
G(p) \approx \sqrt{1 - e^{H_{1/2}(p)}}.
$$

To get some taste of it, here is a plot for the lower, the upper bound and the approximation:

(Sampled for distributions of the form $p(x) \propto (1-x+r)^{-\alpha} + c$, where $r$, $\alpha$ and $c$ are some parameters.)

## Proof

We have:
$$
G(p) = \int_0^1 (2x-1) p(x) dx = \tfrac{1}{2} \int_0^1 \int_0^1 |p(x) - p(y)|dx dy.
$$
(Actually, the right hand side is more general, as it does not require any ordering of p(x).)

For the lower bound we use
$$
|p(x) - p(y)| = \left|\sqrt{p(x)} + \sqrt{p(y)}\right| \left| \sqrt{p(x)} - \sqrt{p(y)} \right| \geq \left( \sqrt{p(x)} - \sqrt{p(y)} \right)^2
$$
and perform the integration.

For the upper bound we use Schwartz inequality for $\left|\sqrt{p(x)} + \sqrt{p(y)}\right|$ and $\left| \sqrt{p(x)} - \sqrt{p(y)} \right|$.