I think this is a good question. Let me start by saying that the subjective experience of the rocket ship is that it passes the event horizon unscathed. (Well... probably not unscathed due to the spaghetti effect unless it is a super, super, supermassive black hole which has a relatively flat event horizon.) An outside observer would see the rocket ship slowly sinking into the event horizon, edging closer but never passing it.
General Relativity allows non-trivial topology in spacetime. For instance, it's possible that the universe is 'closed' in the sense that there is a finite amount of space in it, like there is a finite surface area on the surface a sphere. Consequently, it is not generally possible to use a single coordinate frame (i.e. a single choice of $(x,y,z,t)$ coordinates in one reference frame) which can map out every point in spacetime.
Edit: You said you're familiar with Special Relativity so here's something extra. The Earth's surface, $S^2$, is locally diffeomorphic to $\mathbb E^2$, and likewise, General Relativity models spacetime as a $4$-dimensional pseudo-Riemannian manifold locally diffeomorphic to Minkowski Space, which is the space of Special Relativity.
In your black hole example, here on Earth with our natural $(x,y,z,t)$ coordinates, anything which passes the event horizon no longer has any sensible $t$ coordinate, because the $t$ coordinate explodes to $\infty$ as an object approaches the event horizon (see below). Another way to look at this situation is that our usual choice of coordinates (Schwarzschild) cannot extend into (or out of) the black hole. The event horizon is a singularity in coordinate choice only, not a physical singularity. Much like how the North and South poles of Earth are coordinate singularities in a lot of the common map projections of Earth. Is there something supernatural happening at the poles? Of course not. It's just because we chose to chart our Earth in a particular way.
The same happens when charting $4D$ spacetime.
Schwarzschild Coordinates (Earthly coordinates)
The coordinates we would generally use are Schwarzschild coordinates. Forget about $(x,y,z)$ and let's just look at the $1$-dimensional line straight from Earth to a distant black hole. The spacial distance along this line is $r$. This is the model we would use for the gravitational field of a single black hole where spacetime is otherwise flat (i.e. no dark energy; no dark matter; no expansion).
Here's what the Schwarzschild coordinates look like as we approach a black hole (taken from George Jaroszkiewicz's lecture notes at the University of Nottingham). The black hole singularity is at $r=0$, and the event horizon at $r=1$. We may consider Earth as being at some $r >> 1$:
The lines are called null geodesics and indicate the path that rays of light would take. As you can see, it takes light an infinite amount of time to approach the event horizon from the outside. In fact it takes everything an infinite amount of time to get to the event horizon. But there's an important semantic distinction here: when I say time, I'm really referring to the time that we here on Earth measure, denoted by the $t$ coordinate. It has nothing to do with the subjective experience of someone who's actually falling into the black hole. Their measure of time is called proper time and denoted $\tau$. The yellow 'light-cones' contain all trajectories that objects with mass could take. Here they are drawn for an infalling mass.
If we were to parametrise the path of the rocket ship with $\tau$ in these coordinates, then there would be a special value $\tau_{critical}$ at which it reaches that vertical asymptote $ct \rightarrow \infty$ and its position in these coordinates becomes completely ill-defined. The rocket ship doesn't spontaneously cease to exist at this point; it's just that the rocket ship hits 'uncharted waters' which our chosen coordinate system cannot comprehend.
Retarded Eddington-Finkelstein Coordinates
But there do exist coordinate systems which can map out the trajectory of a mass falling into a black hole. For example, the retarded Eddington-Finkelstein coordinates. Here's a diagram (again taken from George Jaroszkiewicz's lecture notes at the University of Nottingham) of such coordinates applied to a black hole:
Here's what I want you to take home from this diagram: as the mass falls closer and closer to the event horizon, just before it passes it, follow your eye along the null geodesics leading out from the black hole and toward us at $r \rightarrow \infty$. These lines indicate the path that rays of light follow. No matter how far you advance in our Earthly time coordinate, you are still receiving rays of light from when the rocket ship was falling in-- that moment right before $\tau_{critical}$. Any light emitted by the rocket ship once it passes the event horizon, beyond $\tau_{critical}$, is trapped inside the black hole.
Clasically, nothing can escape a black hole once past the event horizon. Not even information. No matter how much time passes on Earth, we receive information (i.e. light) from the instant just before the rocket passed the event horizon, and none from afterwards.
Here I've plotted our time coordinate $ct$ with respect to the retarded Eddington-Finkelstein coordinates:
Following the contour lines, you can see the path that light rays take to reach us from the black hole. No matter which contour line you look at, (i.e. no matter what $t$ coordinate; no matter how long we sit and wait) the light rays always originate from some point with $r>1$ because the contours do not pass through the critical $r=1$ event horizon. And so we will always see the rocket as it was just before $\tau_{critical}$.
The time that the light takes to travel to the observer is not included. The time an event happens in a reference frame is defined as the time that a clock at the location of the event would show. To tell time in a reference frame, Special Relativity imagines clocks at all relevant locations, and all those clocks properly synchronised within the reference frame. It is important to keep that definition in mind, because it also leads to Relativity of Simultaneity, which many novices don't count with.
Which $\Delta t$ is to be used to calculate the time the journey takes for Bob?
In reference frame A, Bob reaches $d_A$ at time $t_{0,A}$ and Alice at time $t_{0,A} + \Delta t_A$, so the journey would take 11 years in reference frame A.
In reference frame B, the journey takes $\Delta t_A /\gamma$ years.
Would Alice see Bobs time run slower or faster? (suppose there is a
clock on board)
As perceived from reference frame A, the clocks in reference frame B run slower. If with "see", you mean what Alice sees through the telescope, then you need to take the Doppler effect into account, and Alice would see the clocks of Bob run faster.
The Doppler shift is caused because each tick of the moving clock occurs at a point a bit closer to Alice than the previous tick. The light from the previous tick had to travel a bit of time to get to that same point, and therefore the two ticks appear closer together to Alice when viewed through the telescope. That is, the clock appears through the telescope to run faster than it actually does.
In an opposite scenario where Bob travels from the earth to the planet, would Alice see Bobs time run slower or faster?
Perceived from reference frame A, Bobs time runs just as slow as on the way towards the Earth. Through the telescope, Alice would see an additional slow-down due to the receding Doppler effect.
Best Answer
Ignoring your first paragraph because magic portals are governed by the whim of fantasy authors, not the laws of physics.
Inertial (velocity-based, non-accelerating) time dilation goes as:
$\Delta t' = \gamma\Delta t = \frac{\Delta t}{\sqrt{1-v^2/c^2}}$
where $\Delta t'$ is the elapsed time measured by a clock in the observed frame and $\Delta t$ is the elapsed time in the observer frame, and $v$ is the velocity that the observer measures between itself and the observed frame.
Qualitatively: the shortest time between two events is always the time measured by the clock for which those events are in the same place, and an observer watching a moving clock always sees the clock advancing slower than the observer's clock.
Gravitational (accelerated frame) time dilation for a non-rotating sphere of mass $M$ goes as:
$\Delta t_f{\sqrt{1-v_e^2/c^2}} = \Delta t$
where $v_e = \sqrt{2GM/r}$, the escape velocity the observer would need to escape the gravity well
$\Delta t$ is the time measured by the observer
and $\Delta t_f$ is the time measured by an observer infinitely far from the massive sphere.
For a circular orbit, in the frame of the orbiting body,
$\Delta t_f{\sqrt{1-\frac {3 v_e^2}{2c^2}}} = \Delta t$
Qualitatively: the shortest time between two events is always measured by the clock whose own escape velocity is greatest, and an observer watching a clock with a lower escape velocity sees the clock advancing faster than the observer's clock, while an observer watching a clock with a higher escape velocity sees the clock advancing slower than the observer's clock.
Both these effects are measurable in everyday life in coordinating clock synchronization between orbiting satellites and ground-based human computers, which is especially important for GPS. The satellite has a relative velocity to the observer on Earth, so we measure its clock running slower. The satellite has a lower escape velocity than the observer on Earth, so we measure its clock running faster. Combining the two gives you the correct clock synchronization. Chart below (wikipedia).