How many electrons are in the d orbitals of NiCl2·6H2O?
The hexahydrate has octahedral symmetry, and it is very probable that this is the case for the solid and dissolved state of the complex.

The compound will not separate into the ions, because the solvation of three ions vs the solvation of one neutral complex is higher in energy, especially considering the fact that opposite charges attract. So in solution state it is a $\ce{[NiCl2(OH2)4]}$ complex with octahedral symmetry.



The $d$-electron count is 8, as it starts with 10 and has a formal oxidation state of II. This is always the case for $\ce{Ni^{II}}$ in any medium but vacuum.

Whether it's a high-spin or low-spin complex depends on the splitting energy of the $e_\text{g}$ and $t_\text{2g}$ molecular orbitals that arise from the $d$ orbitals in combination with the ligand group orbitals, colloquially known as $\Delta_\text{o}$ or $\Delta_\text{oct}$. Have a look at this Wikipedia article on crystal field theory for more details.

In short, there are several factors that influence the magnitude of the splitting:


Oxidation state of central atom
Geometry
Ligand types


The octahedral geometry usually gives good energy splitting.$^\text{[citation needed]}$ In conjunction with a moderate oxidation state compared to Ni(0), and four mid-range splitting ligands (water), it is a safe bet to say that we will have a low-spin octahedral $d^8$ complex: Three pairs of electrons in the $t_\text{2g}$ orbitals and two unpaired electrons in the $e_\text{g}$ orbitals, that is.

Note at this point that the terms high-spin and low-spin are superfluous. The electrons in the $e_\text{g}$ orbitals will always be in a triplet configuration according to Hund's Rule. So it will always be in a "high-spin" configuration from an experimental standpoint.