The problem is that your calculation has no real physical meaning. It is only meaningful to compare two quantities within the same inertial frame. But there is no global inertial frame that connects us to a co-moving galaxy at the Hubble radius: such a galaxy is at rest in its own local cosmological inertial frame (ignoring local peculiar motions). So, due to the Cosmological Principle, its own cosmic proper time is similar to ours.
To obtain meaningful information, we need to analyse the light of that galaxy as we observe it. And as you indicate in your own answer, the Hubble radius is not a cosmological horizon in terms of light paths.
So see why, let us look at the Standard ΛCDM-model in more detail. The expansion of the universe can be expressed in terms of the scale factor $a(t)$ and its derivatives, with $a=1$ today. From the Friedmann equations, it can be shown that the Hubble parameter $H$ has the form
$$
H(a) = \frac{\dot{a}}{a} = H_0\sqrt{\Omega_{R,0}\,a^{-4} + \Omega_{M,0}\,a^{-3} + \Omega_{K,0}\,a^{-2} + \Omega_{\Lambda,0}},
$$
with $H_0$ the present-day Hubble constant, $\Omega_{R,0}, \Omega_{M,0}, \Omega_{\Lambda,0}$ the relative present-day radiation, matter and dark energy density, and $\Omega_{K,0}=1-\Omega_{R,0}- \Omega_{M,0}- \Omega_{\Lambda,0}$. I will assume the values
$$
\begin{gather}
H_0 = 67.3\;\text{km}\,\text{s}^{-1}\text{Mpc}^{-1},\\
\Omega_{R,0} = 0, \quad\Omega_{M,0} = 0.315, \quad\Omega_{\Lambda,0} = 0.685,\quad\Omega_{K,0} = 0.
\end{gather}
$$
It's important to note that, even though the expansion of the universe is currently accelerating ($\ddot{a}>0$), the Hubble parameter $H$ is in fact always decreasing. The cosmic time can then be calculated from
$$
\text{d}t = \frac{\text{d}a}{\dot{a}} = \frac{\text{d}a}{aH(a)},
$$
so that
$$
t(a) = \int_0^a\frac{\text{d}a}{aH(a)}.
$$
Likewise, a photon travels on a null-geodesic
$$
0 = c^2\text{d}t^2 - a^2(t)\text{d}\ell^2,
$$
with $\text{d}\ell$ a co-moving displacement, so that the co-moving distance travelled by a photon is
$$
D_\text{c} = c\int_{a_\text{em}}^{a_\text{ob}}\frac{\text{d}a}{a\dot{a}} = c\int_{a_\text{em}}^{a_\text{ob}}\frac{\text{d}a}{a^2H(a)},
$$
with $a_\text{em} = a(t_\text{em})$ and $a_\text{ob} = a(t_\text{ob})$ the scale-factors at the moments of emission and observation. At any moment $t$, a co-moving distance $D_\text{c}(t)$ can be converted into a proper distance $D(t) = a(t) D_\text{c}(t)$.
The distance to the current Hubble radius is
$$
D_\text{H}=\frac{c}{H_0}\approx 14.5\;\text{Gly}.
$$
Now, we can define two important cosmological horizons: the first is the particle horizon, which marks the edge of the observable universe. It is the maximum distance that light has been able to travel to us between $t=0$ and the present day, i.e. it is as far as we can see:
$$
D_\text{ph} = c\int_0^1\frac{\text{d}a}{a^2H(a)}.
$$
Since $H(a) < a^{-2}H_0\ $ for $a<1$, we get
$$
D_\text{ph} > \frac{c}{H_0}\int_0^1\text{d}a = D_\text{H},
$$
it follows that the Hubble radius is smaller than the particle horizon, and thus part of the observable universe. In fact, $D_\text{ph}\approx 46.2\;\text{Gly}$.
The second horizon is the event horizon. It is the region of space from which a photon that is emitted 'today' will still be able to reach us at some point in the future. Thus
$$
D_\text{eh} = c\int_1^\infty\frac{\text{d}a}{a^2H(a)}.
$$
Since $H(a) < H_0\ $ for $a>1$, we get
$$
D_\text{eh} > \frac{c}{H_0}\int_1^\infty\frac{\text{d}a}{a^2} = D_\text{H},
$$
so the Hubble radius is also smaller than the event horizon: a galaxy on the Hubble radius can still send signals to us (or we to them). $D_\text{eh}\approx 16.7\;\text{Gly}$.
The situation is displayed in these graphs (click on 'view image' for a larger version): the first is with co-moving coordinates, the second with proper coordinates.


The black solid lines indicate our present position. The blue line is the particle horizon through time, the red line is the event horizon, the green area is the Hubble sphere. The dotted black line is a co-moving galaxy that is currently located on the Hubble radius. Its photons that we observe today have travelled on the cyan path (they were emitted at $t=4.3\;\text{Gy}$). Its photons that it emits today ($t=13.8\;\text{Gy}$) will travel on the purple path (they will reach us at $t=49\;\text{Gy}$).
So can we say anything about time dilation? The answer is yes. The light that we observe is also redshifted
$$
1 + z = \frac{\lambda_\text{ob}}{\lambda_\text{em}} = \frac{\nu_\text{em}}{\nu_\text{ob}}.
$$
Within a small time $\delta t_\text{em}$ the source emits a light wave with $\nu_\text{em}\delta
t_\text{em}$ oscillations. Those same oscillations are observed within time $\delta t_\text{ob}$ with frequency $\nu_\text{ob}$, in other words $\nu_\text{em}\delta
t_\text{em} = \nu_\text{ob}\delta t_\text{ob}$, so that
$$
1 + z = \frac{\delta t_\text{ob}}{\delta t_\text{em}}.
$$
In other words, cosmic redshift is directly related to time dilation. Also, within a small time $\delta t_\text{em}$, the distance that light has to travel doesn't change:
$$
\int_{t_\text{em}}^{t_\text{ob}}\frac{c\,\text{d} t}{a(t)} =
\int_{t_\text{em} + \delta t_\text{em}}^{t_\text{ob} + \delta t_\text{ob}}\frac{c\,\text{d} t}{a(t)},
$$
or
$$
\int_{t_\text{ob}}^{t_\text{ob} + \delta t_\text{ob}}\frac{c\,\text{d} t}{a(t)} =
\int_{t_\text{em}}^{t_\text{em} + \delta t_\text{em}}\frac{c\,\text{d} t}{a(t)}.
$$
in these small intervals, the integrands remain constant, so that
$$
\frac{\delta t_\text{ob}}{a(t_\text{ob})} = \frac{\delta t_\text{em}}{a(t_\text{em})}
$$
and
$$
1 + z = \frac{a(t_\text{ob})}{a(t_\text{em})}.
$$
The light from the co-moving galaxy at the current Hubble radius that we observe today was emitted when $a(t_\text{em})=0.403$, so that $z=1.48$, and the observed events from this galaxy are time-dilated by a factor $2.48$. And the light it emits today will be observed when $a(t_\text{ob})=8.07$, with redshift $z=7.07$.
Such time dilations have indeed been observed: we literally see distant supernovae explode in slow motion!