$\ce{NH3}$ and $\ce{H2O}$ are strong and weak field ligands respectively according to spectrochemical series.
No, this is not correct. I keep saying this time and time again. Ammonia is a medium-field ligand at best, and water is the defined ‘zero point’ below which weak field ligands begin.
I understand that pairing of electrons occur in the case of strong field ligands and it does not occur in the case of weak field ligands.
This is simplified at best. The transition from low spin to high spin depends on both the metal and the ligands (and also the geometry). In fact, some complexes are known both in a high-spin and in a low-spin state with identical central metal and ligands. You always need to consider the big picture to determine whether a high spin or low spin complex will result.
I am also going to completely ignore the stuff you wrote about hybridisation. You should not consider hybridisation in transition metal complexes as I have posted numerous times.
To answer this question you should ask yourself how many ligands you have around the nickel centre and which geometry that implies. The complex being tetraamminediaquanickel(II) results in six ligands and thus an octahedral configuration. The next step should be to consider the molecular orbital scheme of an octahedral complex (although the d-orbitals alone carry all the necessary information so even the crystal field model can answer it).
Figure 1: simplified MO scheme of an octahedral $\ce{[ML6]}$ complex ignoring π $\ce{L\bond{->}M}$ interactions. Originally taken from Professor Klüfers’ web scriptum of his coordination chemistry course at the LMU Munich.
The important bit to note is that the d orbitals split into a $\mathrm{t_{2g}}$ and an $\mathrm{e_g}$ part. $\mathrm{e_g}$ corresponds to the $\mathrm{d}_{z^2}$ and $\mathrm{d}_{x^2- y^2}$ orbitals. Thus, a perfect octahedral geometry will generate two degenerate orbitals which should be filled with parallel spins by Hund’s rule. Even if you start to assume Jahn-Teller distortions here (which are not likely, since a high-spin $\mathrm{d^8}$ octahedral configuration is not susceptible to Jahn Teller because there are no unevenly populated degenerate orbitals), that would not lead to an energy difference large enough to induce spin pairing. Remember that a typical 3d-metal octahedral complex is high-spin even in $\mathrm{d^6}$ cases; the energy difference between $\mathrm{t_{2g}}$ and $\mathrm{e_g}$ is generally much larger than what can be achieved by Jahn-Teller distortion.
Now you may decide to argue using square planar complexes. However, the complex in question cannot be square planar since there are six ligands.
Therefore, a magnetic moment of zero is not an option in the case presented; $\ce{[Ni(H2O)2(NH3)4]^2+}$ cannot reasonably adopt a low-spin configuration. The high-spin complex is paramagnetic and in a triplet state ($2S+1 = 3$).
