Not to put too fine a point on it but the big difference between negative energy particles and antiparticles is that negative energy particles have negative energy, that is, for some state $\Omega$ of negative energy,
$$\langle \hat{H} \rangle_\Omega < 0$$
Or, to be more specific since we're doing general relativity, consider the operator of the stress-energy tensor $T_{\mu\nu}$, then, for a null vector $k$,
$$\langle \hat{T}_{\mu\nu}(x)\rangle_\Omega k^\mu k^\nu < 0$$
which has the benefit of being Lorentz invariant.
On the flipside, for reasonable quantum fields, antiparticles have a positive energy. Consider for instance the usual case of a Dirac field. The Hamiltonian (density) operator for it is (in momentum space)
$$\hat{H} = \sum_s \vec{p} (\hat{a}^{s\dagger}_{\vec{p}} \hat{a}^s_{\vec{p}} + \hat{b}^{s\dagger}_{\vec{p}} \hat{b}^s_{\vec{p}})$$
$a^\dagger$ the creation operator for fermions and $b^\dagger$ for antifermions. You can observe that the role of particles and antiparticles is symmetric in the Hamiltonian : any particle will have the same energy as an antiparticle.
On the other hand, consider the usual scalar field, with field operator defined as
$$\phi(x) = \sum_k f_k(x) \hat{a}_k + f^*_k(x) \hat{a}^\dagger$$
with $f_k$ the usual modes $f_k \propto e^{ik_\mu x^\mu}$. The (renormalized) stress-energy tensor, adapted from the classical theory, is
$$\langle \hat{T}_{\mu\nu} \rangle_\Omega = \sum_n (2n |c_n|^2 T_{\mu\nu}[f_k, f_k^*] + n^{1/2} (n-1)^{1/2} c_n c_{n-2}^*T_{\mu\nu}[f_k, f_k] + n^{1/2} (n-1)^{1/2} c_n^* c_{n-2}T_{\mu\nu}[f^*_k, f^*_k])$$
with
$$T_{\mu\nu}[g, h] = (\partial_\mu g)(\partial_\nu h) - \frac{1}{2} \eta_{\mu\nu} (\partial_\sigma)(\partial^\sigma h)$$
and $|\Omega\rangle = \sum c_n |n\rangle$. Then take for instance the state
$$\frac{1}{\sqrt{1 + \varepsilon^2}}(|0\rangle + \varepsilon |2\rangle)$$
Then
$$\langle \hat{T}_{\mu\nu} \rangle_\Omega = (k_\mu k_\nu - \frac12 \eta_{\mu\nu} k_\sigma k^\sigma ) \frac{\varepsilon}{1 + \varepsilon^2} (2 \varepsilon - \sqrt{2} \cos(2 k_\rho x^\rho))$$
The sign of which depends on the last factor. For $\varepsilon$ small enough, there are spacetime regions for which the energy becomes negative.
