One of the nice things about relativity is that it doesn't matter how fast you are traveling, only how fast you are traveling relative to something else. So what does that mean?
- Moving fuel to the engine - as easy as if the ship was stopped
- Moving inside the ship - this could be hard while you're accelerating or decelerating, and there would be a lot of that if you're getting close to light speed. However, the ship's speed doesn't matter - if you're accelerating at 1G, it will be as hard to move around whether the ship's just leaving orbit or already at 0.9c.
- Communicating - from the perspective of anyone/anything on the ship, everything is as if the ship is moving slowly. So signals from the drive can make it to the bridge of the ship just as fast as if the ship was stopped.
- Turning - it's as complicated as if you were turning while in orbit around a planet.
So what is a problem with near-light-speed travel? Reacting to outside stimuli, and getting to and from your desired travel velocity.
Reacting to outside stimuli is dodging, which is not the focus of your question, and navigating. Navigating shouldn't be a significant problem though - if you have the tech to get anywhere near light speed, you should be able to plot your trajectory well enough to make navigating fairly simple.
So getting to and from your desired travel velocity is the only real problem you have to worry about. What's so hard about this? At relativistic speeds, your kinetic energy is described by the following formula:
$$K=mc^2\Big(\frac{1}{\sqrt{1-v^2/c^2}}-1\Big)$$
If $v=\frac{\sqrt{3}}{4}c\approx 0.866c$, which you might not count as a "very high fraction of $c$", we get $K=mc^2=E$. In other words, the ship has as much kinetic energy as energy you could get from converting the entire mass of the ship into energy. So one way to get going that fast would be to have a fuel tank carrying as much mass as the entire rest of the ship, and have a way to convert every atom of fuel into pure energy with 100% efficiency.
With a nuclear reactor, we can currently use about 0.1% of uranium's mass worth of energy. So with a nuclear reactor that could run in space and convert the energy it produces into velocity with 100% efficiency, you could get your kinetic energy up to $K=0.001mc^2$. That gets you up to about $0.045c$. Again, this is if your fuel tank carries as much mass as the entire rest of the ship.
Oh, and don't forget that you have to decelerate once you get to your destination. So "your ship" that you have to accelerate consists of your actual ship and the fuel tank carrying enough fuel to decelerate.
In summary, to get a 10000 metric ton spaceship to $0.045c$, you need 10000 metric tons of uranium to decelerate, and 20000 metric tons of uranium to accelerate. Oh, and a 100% efficient reactor and engine, neither of which can actually exist due to entropy always taking a share. Also you may have realized that if you have 10000 metric tons of spaceship and 10000 metric tons of uranium, you're got to accelerate all the unused uranium.
So getting to a "very high fraction of $c$" just isn't feasible unless you're willing to use up a ridiculous amount of fuel. Unless you consider $0.01c$ to be a very high fraction.
Let's work out an example of an actual "very high fraction of $c$", using an Ohio-class submarine as our starting point. It has a length of 170m and mass of 18750 tonnes. If we scale that lengthwise by 20 we get 3.4km, in the range of your "several kilometers". If we double the cross-sectional radius to make it at least a little bit more cozy, we've now increased the total volume by factor of 80, for an approximate mass of 1.5 million tonnes. I don't know how this compares to what you had in mind for your spaceship, but this is a very long but narrow spaceship. Now how much energy does it require to get this to "a very high fraction of $c$"? To get this ship to $0.99975c$, you need around to convert almost $6*10^{24}kg$ of mass into kinetic energy for the ship. That's the mass of the Earth.
In short, if you're not working with unobtanium/applied phlebotinum/magic you're not likely going to be able to get close to $c$.