Can a bullet really fly through space forever? Some people says that if it would be possible to shoot a bullet so high that it would get over the Earth gravitational pull, the bullet could fly through  space forever, because of no deceleration of friction (and if there wouldn't be any more particles in the space, of course).
But according to quantum physics, every single particle in the universe has a non-zero probability of existence anywhere in the universe. So gravity (produced by gravitons) attract on infinite distance - so there is no place in the universe (even in universe without the particles except Earth) without gravity.
So is it true that bullet would have to stop because of the gravity (even minimal we don't consider the expansion of the universe and curve of the universe)?
 A: 

Can a bullet really fly through space forever?


Essentially, yes. That's kind of what Earth does.


Some people says that if it would be possible to shoot a bullet so high that it would get over the Earth gravitational pull, the bullet could fly through space forever, because of no deceleration of friction (and if there wouldn't be any more particles in the space, of course).


Assuming you had escape velocity after you left the atmosphere, that's about it. There is always friction, but it gets pretty negligible. A bigger concern would be hitting another celestial body as we are letting this experiment go forever.


But according to quantum physics, every single particle in the universe has a non-zero probability of existence anywhere in the universe. 


Equally absurd, one might also ask why you and I haven't popped in and out of existence somewhere else in the universe. The particles of the bullet are atoms, and they are  largely staying together - there are like $10^{23}$ of them. Also, the wave functions decrease exponentially with distance, so for any practical purposes, a centimeter is infinity at atomic scales.


So gravity (produced by gravitons) attract on infinite distance - so there is no place in the universe (even in universe without the particles except Earth) without gravity.


Gravitons are hypothesized, not confirmed, so I won't comment on that, but there is no place in the known universe without gravity, yes.


So is it true that bullet would have to stop because of the gravity (even minimal we don't consider the expansion of the universe and curve of the universe)?


No. Escape velocity is when there is more kinetic energy than the potential energy of the gravitational body.
A: In short:
There is little gas for the bullet to hit - enough to slow it down, but not enough to kill its speed completely; and the probability of being sucked in by the gravity field of a star that it encounters along the way is very, very small.
Full answer:
The density of "space" is about 50 (hydrogen) atoms per cubic meter. According to Wolfram Alpha, the "size of the universe" is about $10^{27}m$, so the total mass of a column that's $1 cm^2$ all the way to the end of the universe (roughly all the matter that the bullet would encounter) is $$50 \cdot 1.6 \cdot 10^{-27} \cdot 10^{-4} \cdot 10^{27}\text{kg} = 8 \text{ gram}$$ - about the mass of the bullet from an AK47. This means all that interstellar gas will slow down your bullet to about half its initial velocity, but it won't stop it.
As for gravity - if you aim carefully, the bullet should be able to evade every massive object that might get in its way. The key here is to realize that a massive object with mass $M$ and radius $r$ has a certain larger "capture radius" $R$ which is a function of $M$, $r$ and $v$ - the velocity of the object trying to get past.
Let's try to compute this - how far from an object would I have to "aim" in order not to be sucked in by its gravity. Note this is a purely classical derivation - I'm not even considering the fact that light curves a bit near a massive object.
From conservation of angular momentum we know that for a velocity "infinitely far away" of $v$, aiming at the edge of the capture disk $R$, the angular momentum will be the same as at the surface where we have a new velocity $v'$ because of gravitational acceleration:
$$v R = v' r\tag1$$
With this velocity we must be going fast enough to overcome gravitational pull, so
$$\frac{mv'^2}{r} \ge \frac{GMm}{r^2}\\
v' \ge \sqrt{\frac{GM}{r}}\tag2$$
Eliminating $v'$ and solving for $R$ (using equality to find the limit):
$$R = r\frac{\sqrt{\frac{GM}{r}}}{v}$$
So the question really becomes: when you point your bullet in an arbitrary direction, will you hit something? In particular, what does that factor $\sqrt{\frac{GM}{r}}$ mean - how big does it get?
According to Wolfram Alpha, the heaviest star in the universe is R136a1, with a mass around 5E32 kg, or 265 times the mass of the sun. For such an object, and a bullet traveling at 700 m/s (AK47 muzzle velocity) the apparent capture radius is
$$R = 4000 r$$
So these heavy objects "look" much bigger than the little dots they are in the sky. It is a bit beyond me to compute what the sky would look like if all stars were that much bigger - would you still have a clear shot across the universe? There are about $10^{23}$ stars in the universe; if you assume they are evenly distributed across all of space, they are about 1.5E19 m apart. From here to the edge of the universe you would encounter about 35 million of them. They are randomly distributed, but let's assume they are just "capture disks" on a series of spheres - and let's compute the fill factor of such a sphere.
The radius of the sun is about 7E8 m, so the "disk" of the sun has a radius of $4000\cdot 7 \cdot 10^8 m$ (I am using the factor 4000 to get a "high" estimate. For most stars, this factor will be smaller). For a "celestial sphere" on which the stars with capture radius R are separated by distance d=1.5E19m, the fill factor is then approximated by
$$FF = \frac{\pi R^2}{\frac{\sqrt{3}}{2}d^2}=1.3\cdot 10^{-13}$$
So the chance of passing one such "shell" is very good. And even passing 35 million is not a problem - if we say probability of capture is $p$ and we have $n$ opportunities, then the probability of escape is 
$$P=(1-p)^n\approx 1-pn$$
In this case, $p=1.3\cdot 10^{-13}$ and $n=3.5\cdot 10^7$.
Fire away. Your bullet will live forever. It has about a 4 in a million chance of not making it to the end of the universe - if the universe wasn't expanding. Taking that into account, it will simply fly on until the end of time.
A: I think your confusion is about whether a bullet could truly escape the earth's gravity if gravity extends forever. You are worried that the bullet will always feel a non-zero gravitational force, and therefore it will stop at some point. 
You are correct that the bullet will feel a non-zero gravitational force. This can be seen from newtonian gravitation; it is not necessary to discuss quantum mechanics. However, as the bullet gets farther and farther from the earth the force gets smaller and smaller. The faster the bullet goes, the faster the force gets smaller. If the bullet is going fast enough, it can happen that the force gets too small too fast, and it cannot stop the bullet, and the bullet will continue forever. The minimum speed the bullet needs to go in order to beat out gravity is called the "escape velocity".
So the moral of the story is that even though the bullet always feels the gravity of earth, if it is going fast enough, the gravitational force will quickly become very weak and the bullet will keep moving away from the earth forever.
