Yes. As long as the universe is decelerating in its expansion, the light emitted from a galaxy beyond the Hubble sphere will eventually reach Earth.
Via Hubble's law, $d_H=c/H$, we see there is this interesting distance at which objects appear to recede from Earth at the speed of light. It exists no matter the expansion rate: inflationary or otherwise. It’s called the Hubble distance, or Hubble radius, or Hubble scale. It is the radius of the Hubble sphere, bounding a region of space with Earth at the center. It might be supposed that it marks the edge of the observable universe — that it is the farthest distance out into the universe that we can presently see. After all, how do distant objects moving at speeds surpassing that of light emit light that has any hope of reaching us here on Earth?
Consider a galaxy located beyond the Hubble radius at a distance greater than $d_H$ that emits a photon towards Earth. Of course, locally this photon is traveling at $v_{pec}=c$ in accordance with special relativity (here, $v_{pec}$ is the peculiar velocity, that is, the velocity of an object measured relative to the expansion). But, on account of the expansion, the photon is initially moving away from Earth with a speed $v_{tot}=v_{rec}−c>0$ (where positive velocities point away from Earth, in the direction of expansion.) Because this galaxy and the light it emits are being swept away from us by the expansion of space, it would indeed seem like this galaxy is forever unobservable. But that would be wrong. Many people make this mistake! The key is to examine how $v_{rec}$ evolves in time. If $H$ is getting smaller in time — if the universe is slowing down in its expansion — then $v_{rec}$ must decrease too because the recession velocity is due to the expansion. And, in the spirit of Hubble’s Law, the farther away the object, the more rapid its deceleration!
We can see this by obtaining an expression for $\dot{v}_{rec}$ when $H$ is allowed to vary: $\dot{v}_{rec}=\dot{H}r+\dot{r}H$, where $\dot{H}$ is the rate of change of the expansion rate. With $\dot{r}=v_{rec}=Hr$, this becomes $\dot{v}_{rec}=r(\dot{H}+H^2)$. If we introduce something called the deceleration parameter, $q=−(1+\dot{H}/H^2)$, then we get something very much like Hubble’s Law, but for $\dot{v}_{rec}$
$$\dot{v}_{rec}=−H^2qr$$ When the universe is decelerating, $q>0$, all objects decelerate: $\dot{v}_{rec}<0$, in proportion to $r$. So, what’s the big deal about $v_{rec}$ getting smaller in a universe that is slowing down? Well, this means that the photon emitted by the galaxy with $r>d_H$, which we said above was moving away from us with a velocity $v_{rec}−c$, is slowing down relative to Earth. When $v_{rec}$ drops to $c$, the photon momentarily is moving neither toward nor away from us (it is locally moving at $v_{pec}=c$ towards Earth but space is carrying it away at $v_{rec}=c$, so $v_{tot}=c−c=0$). Once the recession velocity drops below $c$, the photon begins moving towards us at an ever increasing rate (since now $v_{tot}=v_{rec}−c<0$ and $v_{rec}$ is continuously decreasing making $v_{tot}$ more and more negative). Eventually it reaches Earth, $v_{rec}$ goes to zero, and the photon whizzes into our telescope at the speed of light, $c$! Meanwhile, the emitting galaxy could be well beyond the Hubble radius, receding at superluminal speeds. So superluminal recession of a light source is not, in itself, an impediment to our being able to observe this light. If the universe is decelerating, $d_H$ does not measure the size of the observable universe simply because all things must slow down.
