Okay, trying my luck with a physics answer. Let's first look at the boundary conditions given in the movie, since we're particularly talking about that here. The water planet is said to have $130\%$ of earth's gravitational acceleration on the surface. So we have
\begin{equation}
g_W = 1.3 g_E
\end{equation}
This is a given and not to be violated. And in fact it poses constraints on the relation between both planets masses, radii and densities. With the fact that the planet's volume (a supposed sphere for the sake of simplicity) is $\frac{4}{3}\pi r^3$ We can thus express the planet's radius as a function of its density and its gravitational acceleration:
\begin{equation}
r = \frac{3g}{4\pi G\rho} \quad\sim\quad \frac{g}{\rho}
\end{equation}
We can then fill this into the formula for the escape velocity (and drop some constants):
\begin{equation}
v = \sqrt{2gr} = \sqrt{\frac{6g^2}{4\pi G\rho}} = \sqrt{\frac{3}{2\pi G}}\frac{g}{\sqrt{\rho}} \quad\sim\quad\frac{g}{\sqrt{\rho}}
\end{equation}
So now lets look at the relation between the escape velocities. We want the planet's escape velocity to be lower than that of earth, so:
\begin{align}
v_W &< v_E \\
\frac{g_W}{\sqrt{\rho_W}} &< \frac{g_E}{\sqrt{\rho_E}} \\
\sqrt{\rho_W} &> \frac{g_W}{g_E}\sqrt{\rho_E} \\
\rho_W &> 1.69 \rho_E
\end{align}
So to have a lower escape velocity than earth, the planet would have to have more than $169\%$ of earth's average density.
But in fact, Kip Thorne actually gives an estimate of the planet's average density (in the Technical Notes of his book The Science of Interstellar), namely $10,000 ~\mathrm{kg/ m^3}$, which is indeed $181\%$ of earth's $5,515 ~\mathrm{kg/ m^3}\;.$ Since this is the only actual information we can rely on (and is totally independent of how much water there is on the surface) we can indeed conclude that the escape velocity of Miller's planet is lower than that of earth.
More exactly, the planet's escape velocity would be $\approx 10.8 ~\mathrm{kg/ m^3}$ compared to earth's $\approx 11.2 ~\mathrm{kg/ m^3}\;.$
Best Answer
The gravity from the black hole (BH) will have no effect on their ability to take off from the water planet itself. Objects in orbit feel weightless (think astronauts in the ISS). If you're only worried about getting off the water planet then there should be no problem.
However, if they were to try to put some distance between themselves and the BH then they'd find it a much more difficult task.
Assuming the BH isn't rotating very fast, the time dilation factor for a body in orbit relative to a stationary observer at infinity is:
$$\frac{d \tau}{dt} = \sqrt{1-\frac{3GM}{c^2 r}}$$
The escape velocity of a BH looks the same as it does in Newtonian mechanics:
$$v_0^2 = \frac{2GM}{r}$$
So if the mass of the BH is ~100 million solar masses, the radius at which the time dilation is 60,000 times normal is at $r \approx$ 275 million miles $\approx$ 3x average Earth-Sun distance. The event horizon itself is at $r \approx$ 2x the average Earth-Sun distance. Meanwhile the escape velocity at the planet radius is about 82% the speed of light, or $v_0 \approx$ 250 million meters/second.