A few questions about the concept of work From Wikipedia: The work done by a constant force of magnitude F on a point that moves a displacement d in the direction of the force is the product:
$$W = Fd.$$
If I lift some object from a ground, the force to be put in above equation is the gravitational force $mg$. 
But while I am moving the object upwards, against the gravity, I must pull the object with greater force than $mg$, right? So the net force is $F_{myforce}-mg$ which results in object being lifted. But this is not the case. Why? Does it matter how strongly I pull the object?
What work will I do if I move this object in absence of gravity (in space)? Will work done be zero or not?
 A: You are getting confused between work done by a single force and total work done on a body. When you substitute $F=mg$ in the above equation[1], you are calculating the work done by the gravitational force, only the gravitational force and nothing else.
Substituting $F=F_{my force}$ will give you work done by you on the object. So if $F_{myforce}>mg$, the total work done(sum of the work done by all forces) on the body will be positive, which will show as the body's kinetic energy(velocity).
So from this, you can deduce that if you pull an object in zero gravity, you will do work, on the body. It will just be more than the work you would do on earth, ofcourse because there's no gravity.
[1]A more correct equation for work done would be $W= \vec F\cdot\vec s$, which involves scalar product of two vectors force($\vec F$) and displacement($\vec s$), because you have to take into account the signs(Work done can be negative too).
A: When you accelerate the object the applied force is greated than gravity. The formula is
$$ F = m ( g + \ddot{y} ) $$
where $\ddot{y}$ is the upwards acceleration. The work put in the system is equal to the work needed to lift the object (potential energy) plus the work needed to move the object (kinetic energy).
The work done is $F\, \Delta y$, the potential energy is $m g\, \Delta y$. If the final kinetic energy is zero (object rests after being lifted) then ${\rm Work} = F\, \Delta y = m g\,\Delta y$ indeed. If not at rest in the end, then the work will be
$$ {\rm Work} = F\, \Delta y = m g\;\Delta y + \frac{1}{2} m \dot{y}^2 $$
Without gravity take the above and set $g=0$.
A: Escape for a moment from the definition "work is equal to the force times the distance ('times' meaning scalar product of course)”
The meaning of "work"  may be interpreted as a transfer of energy, or specifically, the transfer in or out of a system's kinetic energy.$^1$   
Your Question: 

If I lift some object from a ground, the force to be put in above equation is the gravitational force mg 

Not necessarily. You could insert any force or combinations of forces that are acting on the system. You just need to understand what is being said by the equation for each case. This is where understanding "work" in terms energy transfer may help. The force you exert lifting the object is transferring energy to the object/Earth system kinetic energy while the gravitational force (mg) is draining the kinetic energy from the object/Earth system. 
If you need to know how to tell if the interaction is adding/subtracting from the system's kinetic energy: If the direction of the motion and the direction of the force (or force component) is in the same direction, then it adds to the kinetic energy, if they're oppositely aligned then the kinetic energy is drained. A little more technically, if the scalar product of the force and distance is positive/negative, then the change in kinetic energy is positive/negative respectively. 

But while I am moving the object upwards, against the gravity, I must pull the object with greater force than mg, right? 

Not correct: you can move the object at constant speed by keeping the gravitational force and the upward lifting force balanced. When the object is on the ground you will need to exceed this force to accelerate the object to lifting speed, and then lift with less force than the weight to slow it down to rest. Again, in the energy picture, if the net flow of energy is zero (forces balance) the kinetic energy will be a constant.$^2$ 

What work will I do if I move this object in absence of gravity (in space)? Will work done be zero or not?

If you apply a force on the object over some distance then work will be done. In other words, if you change the object's kinetic energy then work will be done.
Final note: I am not advocating an energy description over the force description. Being able to switch between perspectives will add to depth of understanding the physics involved.
$^1$ If the transfer is of a microscopic nature or caused by a temperature difference, then stick to using "heat" to describe the energy transfer.
$^2$ It's actually the net energy current, a.k.a. net power, that has to be zero to keep the speed constant. You could have a net flow of zero and have a period of speeding up followed by a period of slowing down in such a way as to keep the average speed constant, but not the instantaneous speed everywhere constant.
