Second quantization with qubits Is "second quantization" means system wich can contain variable, unknown, superposed and otherwise uncertain number of qubits?
Can "second quantized" system contain 0.5% of 1 qubit and 95% of 2 qubits?
Does this mean that quantum field state cannot be described with quantum computer with fixed number of qubits?
Or may be this is wrong and the "power" of qubit is sufficient?
 A: In general q-bits can not be mapped to fermion or bosonic fields, but in controlled cases they can.
As you are arguing, the size of the Hilbert space is the main constraint.  Since q-bits have a finite dimensional Hilbert space,  the fermions and bosons must live on a finite sized lattice and the number of particles must be restricted.
The next difficulty is mapping the fermion or boson operators to q-bit operators.
Boson operators are easiest.  If you fix the number of particles on a site to $N=2^m$, then you need m q-bits per site.  The binary number given by this q-bit register tells you how many number of bosons are on that site.  The creation operator/annihilation operator will have to be implemented with a sequence of gates acting on this q-bit register.  The nice thing is that bosons commute, so adding sites, is as simple as adding registers.
Fermions are nicer in that each fermion mode maps directly to a q-bit, since each fermion mode can only be occupied or unoccupied, but to get the right stastics from gate opeartions, you need apply the jordan wigner transformation to go from spins(q-bits) to fermions or vise versa.  For 1D, This is simple, but for higher dimensions, it's more complicated. see https://en.wikipedia.org/wiki/Jordan%E2%80%93Wigner_transformation
