Does the Hubble constant depend on redshift?
Yes, definitely.

The Hubble constant describes the expansion rate of the Universe, and the expansion may, in turn, may be decelerated by "regular" matter/energy, and accelerated by dark energy.

It's more or less the norm to use the term Hubble constant $H_0$ for the value today, and Hubble parameter $H(t)$ or $H(a)$ for the value at a time $t$ or, equivalently, a scale factor $a = 1/(1+z)$, where $z$ is the redshift.

The value is given by the Friedmann equation:
$$
\frac{H^2(a)}{H_0^2} = 
\frac{\Omega_\mathrm{r}}{a^4} +
\frac{\Omega_\mathrm{M}}{a^3} +
\frac{\Omega_k}{a^2} +
      \Omega_\Lambda,
$$
where 
$\{
\Omega_\mathrm{r},
\Omega_\mathrm{M},
\Omega_k,
\Omega_\Lambda
\} \simeq
\{
10^{-3},0.3,0,0.7
\}
$
are the fractional energy densities in radiation, matter, curvature, and dark energy, respectively.

For instance, you can solve the above equation at $z=0.1$ and find that the expansion rate was 5% higher than today.

Since everything but dark energy dilutes with increasing $a$, $H(a)$ will asymptotically converge to a value $H_0\sqrt{\Omega_\Lambda} \simeq 56\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$.

The figure below shows the evolution of the Hubble parameter with time:



As noted by KenG, the fact that $H$ decreases with time may seem at odds with the accelerated expansion of the Universe. But $H$ describes how fast a point in space at a given distance recedes. Later, that point will be farther away, and so will recede faster. From the definition of the Hubble parameter, $H\equiv\dot{a}/a$, multiplying by the scale factor shows the acceleration $da/dt$: