If we free a metal plate of free electrons will it become weaker to external mechanic forces? If we free a metal plate of free electrons by means of a strong electric field will it become then weaker in case of applied external forces? To make it simpler, do free electrons have any role in connections between lattice atoms?

 A: The free electrons in a metal are not really free. What we really mean when we call them free is that they do not completely fill the conduction band and thus can change their energy in response to a perturbation.
Furthermore, if we go beyond simple band theory and include the Coulomb interactions between the electrons, the so-called free electrons are really quasiparticles of the Landau liquid with properties very different from those of the electrons outside of metal. The thermodynamic properties of such an electron gas can be calculated (see, e.g., the detailed discussion in Fetter&Walecka's book) and they play an essential role as a part of the cohesive energy/force - the force holding the solid together. This is to say that the solid would fall apart if all the free electrons are removed.
Another take on it can be done from the chemistry viewpoint: we are dealing here with a so-called metallic bond, a collective bond between ions mediated by shared electrons. E.g., in a $Na$ crystal, the monovalent $Na$ atoms each share their single electron - these electrons are free to run over the crystal, but they are essentially  a part of the chemical bond. Without these electrons we would have a bunch of positive $Na^+$ ions and the crystal would not hold together.
Remark
The number of electrons displaces between the plates of a capacitor in a typical electric circuit is negligible compared to the total number of atoms and free electrons in the capacitor plates.
Indeed, the total charge of the electrons contained in 1 mol of substance is about $10^5$Coulombs:
The magnitude of the electrical charge of one mole of elementary charges 
(approximately 6.022×1023, the Avogadro number) is known as a faraday unit 
of charge (closely related to the Faraday constant). One faraday equals 
96485.33212... coulombs.[10] In terms of the Avogadro constant (NA), one 
coulomb is equal to approximately 1.036×10−5 mol × NA elementary charges.

This is comparable to a charge of an automobile battery, but many order of magnitude bigger than the charge of a typical capacitor: see here.
A: When one puts a capacitor at the ends of a battery, the metal plates on the negative side will have a lot more charge and the other plate will display a positive charge, which means a lack of electrons. So in this sense you do free  electrons from one side and add them on the other.

do free electrons have any role in connections between lattice atoms?

As long as the currents induced in the circuit while "freeing electrons" are within values that will not start melting the circuit, the plates will be unaffected too.
High currents start melts, i.e. destroying lattices.

A good electrical conductivity is the same as a small electrical resistance. An electric current will flow through all metals, however they still have some resistance, meaning the current needs to be pushed (by a battery) in order to keep flowing. The bigger the resistance, the harder we have to push (and the smaller the current is). Current flows easily through copper thanks to its small electrical resistance, without much loss of energy. This is why copper wires are used in mains cables in houses and underground (although overhead cables tend be aluminium because it is less dense). However, where size rather than weight is important, copper is the best choice. Thick copper strip is used for lightning conductors on tall buildings like church spires. The copper strip has to be thick so that it can carry a large current without melting.

So it will be the process of removing electrons that generates melting currents destroying the solid state lattice.
