When you ask for a "perfect" or "true" inertial reference frame you are asking for something that cannot be answered in physics. Perfection is only possible in mathematics, not physics.
So in physics, what can be asked is whether or not a given reference frame is an inertial frame to a certain level of accuracy. The surface of the earth is not an inertial frame because of the gravitational field of the earth - not because the earth is moving around the sun and the sun is moving around the galaxy. But if you consider motion only in a horizontal plane on the surface of the earth and if you are only doing the typical high school physics tabletop experiments, the earth is an inertial reference frame as far as the accuracy of the measurements performed is concerned. If you do more accurate measurements, then it would not be an acceptable inertial reference frame.
Consider a satellite in orbit around the earth and examine a relatively small volume near the center of mass of the satellite. That small volume over a suitably small period of time will be an inertial reference frame to a very high level of accuracy. For example, two small masses that are 1 inch apart (radially) in orbit around the earth that start out "exactly" at rest relative to each other will over a time period of 10 seconds come to have a relative speed of 0.006 inches/second due to the differences in orbital velocity for two orbits that differ by 1 inch. So it depends on the level of accuracy needed for an experiment that you want to perform in an inertial reference frame.
To get a reference frame that is more accurately inertial it would necessary to be orbiting much further from all gravitating objects. Thus, it is all about the level of accuracy you require of the inertial reference frame.
I've only skimmed the Wikipedia article you link to. From a quick look I'd say the paragraphs you quote are making points about what a theory of gravity needs to look like. For example you say "Curvature of spacetime in only required in order to explain tidal forces", but what that really means is that it's impossible to have a theory of gravity without curvature. That's because any theory of gravity inevitably has to describe tidal forces. You go on to say "as long as you ignore tidal forces, you can explain gravity without curvature", but you can't ignore tidal forces so you can't explain gravity without curvature.
To take your two specific questions:
Question 1. Gravity i.e. General Relativity isn't a theory of forces: it's a theory of curvature. By focussing on the "fictitious forces" you're getting the wrong idea of how GR works. When you solve the Einstein equation you get the geometry (curvature) of space. This predicts the path a freely falling object will take. We call this a geodesic and it's effectively a straight line in a curved spacetime. If you want the object to deviate away from a geodesic then you must apply a force - and there's nothing fictitious about it.
For example, GR predicts that spacetime is curved at the surface of the Earth, and if you and I were to follow geodesics we'd plummet to the core. That we don't do so is evidence that a force is pushing us away from the geodesic, and obviously that's the force between us and the Earth. But, and it's important to be clear about this, the force is not the force of gravity, it's the force between the atoms in us and the atoms in the Earth resisting the free falling motion along a geodesic.
Question 2. Again this is really just terminology. When you're free falling "gravity" is not eliminated. Remember that "gravity" is curvature, and in fact the curvature is the same for all observers regardless of their motion. That's because the curvature tensor is the same in all co-ordinate frames. The existance of tidal forces is proof that gravity/curvature is present.
When you're free falling you are moving along a geodesic. It is true to say that there are no forces acting, but this is always the case when you are moving along a geodesic. Remember a geodesic is a straight line and objects move in a straight line when no forces are acting. There would only be a force if you deviated from the geodesic e.g. by firing a rocket motor.
Response to fiftyeight's comment: this got a bit long to put in a comment so I thought I'd append it to my original answer.
I'm guessing your thinking that if you accelerate a spaceship it changes speed, so when you stop something has happened, but when the Earth accelerates you nothing seems to happen. The Earth can apply a force to your for as long as you want, and you never seem to go anywhere or change speed. Is that a fair interpretation of your comment?
If so, it's because of how you're looking at the situation. Suppose you and I start on the surface of the Earth, but you happen to be above a very deep mine shaft (and in a vacuum so there's no air resistance - hey, it's only a thought experiment :-). You feel no force because you're freely falling along a geodesic (into the Earth), while I feel a force between me and the Earth. From your point of view the force between me and the Earth is indeed accelerating me (at 9.81ms$^{-2}$). If you measure the distance between us you'll find I am accelerating away from you, which is exactly what you'd expect to see when a force is acting. If the force stopped, maybe because I stepping into mineshaft as well, then the acceleration between us would stop, though we'd now be moving at different velocities. This is exactly what you see when you stop accelerating the spaceship.
It's true that a third person standing alongside me doesn't think I'm accelerating anywhere, but that's because they are accelerating at the same rate. It's as though, to use my example of a spaceship, you attach a camera to the spaceship, then decide the rocket motor isn't doing anything because the spaceship doesn't accelerate away from the camera.
Best Answer
The answer by Luboš Motl was describing the contribution of spacetime curvature as deviations from flat spacetime. In flat spacetime, you can always find a global inertial reference frame, and due to the "automatically guaranteed" fact he mentioned we know that in curved spacetime you can always find a local inertial frame where deviations from inertial are second order in space and time.
Flatness means that gravity is uniform or absent. Spacetime curvature means that gravity is non-uniform, which is tidal gravity. The further away you go in curved spacetime the more non-uniform gravity is. This causes spatially separated geodesics to accelerate relative to each other more than nearby geodesics.
Luboš Motl made one mistake in his answer, and that was to claim that the greatest contributions come from the moon and the sun. The largest source of non-uniform gravity near the earth is the earth itself. Due to the curvature and the finite size of the earth, geodesics on opposite sides of the earth accelerate towards each other at 2 g, which swamps any contribution from the moon or sun. It is this non-uniformity in gravity, this curvature of spacetime, that keeps the surface of the earth from expanding although it is accelerating outward at 1 g.
As your region of interest covers a larger and larger area, the deviations from flatness increase. So in a small lab, we can treat free-falling objects as inertial. If they start out at rest with each other then they stay approximately the same distance apart. Over a large region, such as the whole earth, that doesn't work. Free-falling objects initially at rest with each other do not stay the same distance apart.
No. They have nothing to do with the earth's rotation. The effects described above by me and in the other answer by Luboš Motl are for a non-rotating object. Earth's rotation produces an additional but very small effect. I have neglected that here.
Basically, for the curvature corrections to be small you need to have a small enough region of space and time so that objects at rest in free-fall do not appreciably change their distances with each other. If your room is small enough that everything falls in the same direction then you should be fine. If you have a lab that spans a continent then objects falling on one side of the lab will be accelerating slightly towards objects falling on the other side of the lab.