Do cosmological event horizon evaporates like black-hole horizon?

While the inflation phase of the universe, expansion was exponential and so the universe was de Sitter-like. So, for a point-like observer, there was a cosmological event horizon. Do this horizon evaporates due to the emission of particles? If this horizon evaporates does this mean that the expansion tends to be naturally not in a de Sitter phase?

Edit:

What I’ve tried so far:

The temperature of the de Sitter horizon is:
begin{equation}
T=frac{H}{2pi}.
end{equation}
Where $H=frac{dot{a}}{a}$. The area of the horizon is:
begin{equation}
A=frac{4pi}{H^2}.
end{equation}
I don’t know if I can do this but let’s say that the luminosity inside de horizon due to Hawking radiations is $L=sigma A T^4$, where $sigma$ is the Stefan-Boltzmann constant. Assuming that the energy associated to the Horizon is $E_text{dS}=alpha H^2$ ($alpha$ is a proportionality constant), we thus have :
begin{equation}
frac{d H^2}{dt}=-frac{2sigma}{alpha (2pi)^3}H^2.
end{equation}
Whose solution is:
begin{equation}
H(t)=H(0)e^{-frac{sigma}{alpha (2pi)^3}t}
end{equation}
Then, we find an $a$ of the following form:
begin{equation}
a(t)=C,text{exp}left( -frac{alpha (2pi)^3}{sigma}H(0)e^{-frac{sigma}{alpha (2pi)^3}t} right)
end{equation}
Which tends to $0$ at $trightarrow-infty$, is exponential when $tsim0$, and tends to give a Minkowski-like metric for $trightarrow +infty$