If you could make it down to the core, then yes, you would probably be able to experience a "solid surface" (where I put that in quotes for reasons that should be apparent in a moment).
The question really gets to, though, what you consider to be a "solid" and a "surface" in a gas giant. The issue at hand is what the Wikipedia article stated - as you go down through the atmosphere, you encounter denser and denser material. Gas under incredibly high pressure will start to behave like a liquid, and be as dense or denser than a liquid such as water. There would be no definite point at which you could say the stuff above you is clearly "air-like" while the stuff below you is clearly "water-like," it's a gradient.
You would also get crushed long before you made it anywhere near the core, just like it's only fairly recently that we've been able to build submersible vessels that can go to the deepest parts of the ocean on Earth.
They are sometimes also called double planets and they're more widespread in fiction than in observations. I don't think that there is any new instability that would appear for the system of double planets orbiting a star and that wouldn't be present for other, more asymmetric pairs of planets. Obviously, the tidal forces would be really large if the planets were close enough to each other. But because the tidal forces go like $1/r^3$, it's enough to choose the distance that is 5 times larger than the Earth-Moon distance and the tidal forces from the other Earth would be weakened 125 times and would already be as weak as they are actually from the Moon now (with a lower frequency).
One must realize that the systems with two or several planets are rather rare and the condition that the mass of the leading two planets is comparable is even more constraining.
Imagine that each of the two planets has a mass that is uniformly distributed between 1/20 of Earth's mass (like Mercury) and 300 Earth masses (like Jupiter) on the log scale. The interval goes from the minimum to the maximum that is 6,000 times heavier. That's more than 12 doublings, $2^{12}=4,096$. So if you pick the first planet to be at a random place on that interval of masses (uniformly at the log scale), the probability that the second planet's mass (which is independent) differs by less than the factor of $\sqrt{2}$ from the first mass is about $1/12$.
Only $1/12$ of systems that look like a pair of planets will be this symmetric. And the number of pairs of planets - even asymmetric ones – is rather low, indeed. The reason is really that during the violent eras when Solar-like systems were created, rocks had large enough velocities and they flew in pretty random directions so they were unbound at the end. It's just unlikely to find two large rocks in the same small volume of space: compare this statement with some high-temperature, high-entropy configurations of molecules in statistical physics.
It's also rather unlikely that a collision with another object creates two objects that will orbit one another. After all, the two-body orbits are periodic so if the two parts were in contact during the collision, they will collide again after one period (or earlier). Equivalently, the eccentricity of the orbit is likely to be too extreme which will lead to a fast reunification of the two new planets. Moreover, even if something would place the two newly created planets from a "divorce" on a near-circular orbit, perhaps a collision with a second external object (good luck), it's very unlikely that such an orbit will have the right radius, like the 1 million km I was suggesting in the case of the hypothetical double Earth above. If the two objects are too close, the tidal forces will be huge and (at least for some signs of the internal angular momentum) they will gradually make the planets collapse into one object again. And if the energy with which the planets are ejected from one another is too high, no bound state will be created at all. So the initial kinetic energy of the newborn 2 planets would have to be almost exactly tuned to their gravitational potential energy (without the minus sign) and that's generally unlikely, too.
Best Answer
My answer is more of a summary of insight that others have presented on a number of questions on Physics.SE.
It is true that all astronomical bodies larger than a certain mass will take on a nearly spherical shape. The logical process to arrive at this conclusion involves several steps. I will try to enumerate these with the smallest number of non-trivial steps.
I have a hard time understanding what the Star Wars pictures are even trying to depict, but I will focus on Lola Sayu since I think I can make out what the picture is showing. The depiction is akin to an apple with a bite taken out of it.
(Image license CC-BY-SA-3.0, Wikimedia Commons)
Specifically, here are the various reasons such a shape is unphysical for a planet:
As a final note, the pieces blow off from the planet in the picture are either in orbit, or they will be cleared away within a fairly small amount of time. Most objects will probably not remain in orbit since they are ejected from the surface, and it will more than likely return to the surface at some point in the future, per:
(Image license CC-BY-SA-3.0, Wikimedia Commons)
Now, obviously you can't tell if something is in orbit from the picture (since a still picture doesn't show movement), but it leaves plenty of unanswered questions. Where did that mass of the planet go?! Perhaps it blew away faster than escape velocity. Either way, I'm pretty sure none of the concerns mentioned here were given consideration in the creation of the artwork.