I saw three different depictions:
Never ending 'hole':
Point:
Paraboloid:
Also, do black holes rip space-time, if not, is there any other way to?
P.S. I'm not sure if the last one is a point as well.
[Physics] What does spacetime look like near a black hole
black-holesgeneral-relativitygravitysingularitiesspacetime
Related Solutions
It's a difficult question to answer because in relativity, distance and volume are coordinate-dependent quantities. Even without a black hole, special relativity allows us to distort distances (Lorentz contraction) to an arbitrary degree by moving at speeds close to lightspeed.
That being said, if we limit ourselves to Schwartzschild coordinates (which basically represent how an observer hovering at a distance from the black hole sees things), there are some interesting facts to examine.
First, the coordinate distance to the event horizon is indeed greater than would be expected in flat space, meaning if you hovered over the black hole and let down a tape measure, you'd need a longer-than-expected tape measure, by a factor of over 2, to reach the event horizon. This means the volume of the region around the hole (outside the event horizon) would be greater than that contained in a region of flat space with the same circumference.
Diagrams like the ones you posted are basically slices of the spacetime at a constant Schwartzschild time coordinate, so the relationship of distances and volumes they depict is qualitatively accurate. For objects like neutron stars that are close to being a black hole but not quite, there is also significant additional distance and volume, but less than for black holes.
On the other hand, if you freely fall into a black hole then you hit the event horizon and then the singularity in a finite time from your point of view. There are other coordinate systems, namely Lemaître and Gullstrand–Painlevé, that describe the spacetime "from the perspective of falling observers" in some sense. If you measure the volume of a constant-coordinate-time slice in those coordinates, you'll get something time-dependent (for Lemaître) or equivalent to flat space (for G-P). This just goes to show that the "volume of space" in a curved spacetime region really isn't a well-defined quantity.
It depends what you mean by 'Would remain the same'. If you are asking about the exact precise orbit, the answer is yes, it would change, since the Schwarzschild (and also Kerr solution of a rotating black hole) solution is a vacuum solution of the Einstein equations, so it describes strictly speaking only Vacuum, with a curvature singularity in the centre. Since the Sun is made out of matter, has a finite size and you need many, many variables to fully describe it (for example it is non-spherical) the metric around it is not Schwarzschild (remember that a Schwarzschild black hole is only described by one parameter).
However, the Schwarzschild metric is a good approximation around the Sun, and for current observational purposes it describes nature well. What one should do, is to calculate the perturbations around the Schwarzschild metric in some form of power series in the characteristic length scale of the Sun over the Schwarzschild radius of the Sun. You will then find that these corrections are negligible at our current measurement capabilities. This last statement is with the exception of the advancement of the perihelion of Mercury, where we can now measure the quadrupole moment $J_2$ of the sun to be non-zero. This means that one should replace the Sun not with a Schwarzschild black hole, but with a Kerr black hole (to achieve a non-vanishing quadrupole moment).
As an aside, it is also important to note that the parameter that describes a Schwarzschild black hole $M$ is precisely constructed such that the physics around that central object reduces to the orbits of a body with mass $M$ in Newtonian gravity. It is therefore not surprising that the orbits do not change that much far from the event horizon.
Best Answer
These diagrams try to depict a slice of spacetime, but likely confuse it with a gravity well. A gravity well diagram depicts with an altitude for each location how much potential energy a particle in that location would have. There is also likely some confusion with the classic rubber sheet analogy used to describe gravity (plus, in many media, graphical designers not using the right shape to illustrate an article). A spacetime diagram instead tries to depict how spacetime is curved, not how much energy you would have in a particular spot.
The closest we can get to that for a non-rotating black hole is Flamm's paraboloid: $$z(r)=2\sqrt{r_s(r-r_s)}$$ where $r_s$ is the Schwarzschild radius. This is a surface in 3D space that corresponds to a 2D slice of the black hole spacetime in the equatorial plane for a given time. Distances between points measured along the surface correspond to distances measured between points in the black hole spacetime.
However, the paraboloid only represents the spatial curvature and not the temporal curvature. It represents a single "moment in time" and a particle moving in spacetime will not stay on the paraboloid. One could imagine other paraboloids stacked on top of the first one with a distance corresponding to the time distance $dt^2$, but since this varies with radius the picture becomes pretty unclear immediately.
Also, note that this parabolid ends at the event horizon $r_s$. It is however entirely valid to consider the full parabolid $$z(r)^2=4r_s(r-r_s),$$ but the meaning is subtle: the lower half does not correspond to the interior of the black hole, but a so-called Einstein-Rosen bridge which is a form of wormhole.
To really understand the structure of spacetime Penrose diagrams are useful because they actually try to show the topology of (a part of) spacetime. For a black hole they make it clear that the singularity is not pointlike, neither a circle or some other space-like shape, but a space-like thing similar to a particular moment in time. This is why it is so unavoidable and hard to illustrate.