Q:

Would mining huge amounts of resources on the Moon change its orbit?

Imagine that humanity builds mining facilities on the moon. We know that the moon has Helium-3 which could be used as fuel for fusion reactors. Also, there are a lot of other resources which could be used for building ships and buildings instead of transporting them from Earth. 
Thousands of tons of metals will be mined from the moon in order to sustain a giant fleet of military, civilian and scientific ships for further colonization of our Solar System.
Because all of these actions, the moon will start to lose its own mass. Would this have some kind of impact on its orbit? If it is, how severe would the changes be?

A:

The mass of the Moon is 7.342×1022 kg.
One ton is 103 kg. How much is thousands of tons? Let's say you have thousands of thousands of tons. That's one million tons, or 109 kg. This is still ten thousands times a billion less than the mass of the moon.

(source: Diego Delso)
Just as a comparison, one of the largest mines that ever operated on Earth, Chuquicamata in Chile, produced much more than "thousands of tons":

...it remains the mine with by far the largest total production of approximately 29 million tonnes of copper to the end of 2007...

And even though it's a big hole in the ground, it is completely negligible relative to the mass of the Earth (or the Moon for that matter).
To answer your question:

Does mining huge amounts of resources on Moon will change its orbit?

The answer is no.
EDIT - the below is wrong because I don't know physics as much as I thought
Let's assume you mined 100 Chuquicamatas on the Moon and removed this mass to build your colonisation fleet. Let's ignore effects of momentum and potential energies etc. The orbit velocity is defined by $v=\sqrt{GM/r}$, where $M$ is the mass. Assuming constant orbit, the new velocity is defined by $v' = \sqrt{M_0 / M_1}$.
Let's plug in some numbers:
$v' = \sqrt{\frac{7.342×10^{22}}{7.342×10^{22} - 100\times26×10^6}} = 1.0000000000000178$
The Moon's orbit is going to be 1.0000000000000178 times faster.
EDIT - so let's change it a bit.
Since $M$ in the equation $v=\sqrt{GM/r}$ is the mass of Earth, the mass of the Moon means nothing. Then the velocity to radius ratio is fixed regardless. It will not change a thing!
However, we can still calculate the mass change percentage. It will be:
$\frac{7.342×10^{22} - 100\times26×10^6}{7.342×10^{22}} = 0.99999999999996458$
The Moon's new mass will be 99.999999999996458% of the Moon's old mass. Completely meaningless.

Just as an aside, it is often talked about mining asteroids, moons, and all kinds of other extraterrestrial bodies. For example, as you said, the Moon has plenty of helium-3. Asteroids have plenty of precious metals like platinum or iridium. But this is pointless, because you know what place has even more helium, platinum, and iridium? Earth. Earth has much more. And it's easier to mine because you don't need to build spaceships and facilities in hostile environments to do it.
EDIT 2 - space mining
Some comments mentioned that getting things out of the Moon is easier than it is on Earth, and that you can dump things on the Moon without environmental impact. It doesn't work like this.
In films and video games you "mine a resource" (e.g. iridium) from a planetary body. In real life, you build the mining facilities, you build the refining and smelting facilities, you need people to do it, even if you have robots you need people to fix the robots, you need to feed the people, entertain the people, you need a constant supply of consumables to refine the stuff. And this is only the "resource". You don't build spaceships out of helium. You build them out of steel/aluminium/carbon-composites. So you need to mine that as well, and you need to smelt that as well. You also need to build everything on the Moon because otherwise you need to transport all your resources to another place, so you need factories. To do all of that, you are going to need quite a lot of population. And then the environmental factor becomes important. Your mining will generate huge amounts of dust (combination of dryness and low gravity). This just doesn't work.

A:

In order to change the Moon's orbit, you'd have to change its velocity.
You could mine away half its mass without significantly changing its velocity, as long as all the mass was lifted off by something like rockets.  If you use mass drivers, and they all throw the same direction, you'd start to see detectable changes in the Moon's orbit by the time you'd thrown a couple million tons (a tiny fraction of one percent of the Moon's mass, but thrown at Lunar escape velocity).
If you use mass drivers that throw in different directions (perhaps you're sending the product all over the Solar System, rather than all to Earth or LEO), once again, the average impulse could be adjusted to zero, keeping the Moon (the part you haven't thrown into space, at least) right where you found it.

A:

To a first approximation, simply removing mass from an orbiting body doesn't change its orbit. The orbital radius is given by the velocity and the gravitational pull of the primary (in this case, Earth). When the mass of the satellite is small compared to the mass of the primary, the satellite mass cancels out in the equation and is not relevant.
To see a concrete example, what happens when an astronaut makes a spacewalk from the ISS? Nothing - the astronaut and the ISS continue to follow the same orbital path, even though the ISS is hundreds of times more massive than the astronaut.
As others have touched upon, the process of removing vast amounts of mass might have an effect, depending on how it's done:

If you use chemical rockets, for instance, the reaction to the rocket thrust at take-off would have some effect eventually.
An electric rail gun launcher (as imagined by Arthur C. Clarke) would transmit the acceleration force through its structure to the Moon and so deliver a torque.
a space elevator would have no direct effect.

But these are outwith the scope of the question.

