The short answer is special relativity.
classically
According to classical mechanics, there is nothing to prevent an object from accelerating faster than the speed of light. By conservation of energy you can calculate the speed of a dropped object by looking at the change in gravitational potential. As you note, the equation $PE=mgh$ needs to be modified to handle the changing gravitational field $g$ once the object is far from the surface of the Earth.
We can use the Newtonian gravitational potential $V(r) = - \frac{GM}{r}$, where $r$ is the separation between the central mass $M$ and the falling object. The gravitational potential energy is $PE=mV=- \frac{GMm}{r}$. Dropping an object from rest we find:
$$ E_\mathrm{initial} = E_\mathrm{final} $$
$$ -\frac{GMm}{r_\mathrm{initial}} = -\frac{GMm}{r_\mathrm{final}} + \frac{1}{2}\, m\, v^2 $$
$$ v = \sqrt{ 2\, GM \left( \frac{1}{r_\mathrm{final}} - \frac{1}{r_\mathrm{initial}} \right) } $$
By choosing $M$ and the $r$'s appropriately we can make $v$ whatever we want.
special relativity
If we want to use special relativity, we need to modify the expression for kinetic energy. Special relativity also accounts for the rest energy of the falling particle. The rest plus kinetic energy is $E = \gamma\, mc^2$, where
$$\gamma = \frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^2}}$$
is the Lorentz factor for a moving particle. If $v=0$, then $\gamma=1$ so the energy is just the rest energy $mc^2$.
By conservation of energy
$$ -\frac{GMm}{r_\mathrm{initial}} + mc^2 = -\frac{GMm}{r_\mathrm{final}} + \gamma\,mc^2 $$
$$ (\gamma - 1)\,c^2 = GM \left( \frac{1}{r_\mathrm{final}} - \frac{1}{r_\mathrm{initial}} \right)$$
The velocity of the dropped particle is hiding in the $\gamma$. After some algebra we can find:
$$ v = c \sqrt{1 - \left[ \frac{1}{\frac{GM}{c^2} \left( \frac{1}{r_\mathrm{final}} - \frac{1}{r_\mathrm{initial}} \right) + 1} \right]^2}$$
You can check and see that the square root term is always less than 1, so the dropped particle does not go faster than the speed of light!
Better yet, you could use general relativity instead of Newtonian gravity... but that's beyond the scope of this question.
The potential energy has a gauge freedom, that is we can define the zero to be anywhere we want without affecting the physics. A side effect of this is that we cannot experimentally measure potential energy, we can only measure differences in potential energy.
So when you say the potential energy of an object raised to a height $h$ is $mgh$, what this really means is that raising an object by a distance $h$ changes the potential energy by $mgh$ i.e. the difference in the potential energy before and after raising was $mgh$.
The person holding the suitcase can define its potential energy to be zero, but this is just a choice of gauge. Regardless of how the person holding the suitcase defines the potential energy it still changes by $+mgh$ when it is raised by a distance $h$ and $-mgh$ if it is lowered by a distance $h$.
Best Answer
Saying the potential energy of the object is $mgh$ is technically incorrect, but nothing bad happens if you think of it like this, at least for simple systems.
The reason is that we can just think of the object as the system. Then the constant gravitational field is external to the system. If we want to look at changes in mechanical energy of the object moving under the force of gravity only, then we have
$$\Delta E=\Delta K=W_\text{ext}=-mg\Delta h$$
or we can just say that
$$\Delta K+mg\Delta h=0$$
This $mg\Delta h$ is usually stated as the change in potential energy of the object. But it doesn't really matter what you call it as long as you recognize the validity of the above equation.
In general you do need to be careful with it. For example, if we have two planets approaching each other, if we were to add up the (incorrect) changes in "individual potential energy", then we would be double-counting the change in potential energy.