Why is ruthenium(III) intensely colored in water
As you said, in $\ce{RuCl3}$, the charge of the Ruthenium ion is +3, meaning d5. In water, the RuCl3 becomes $\ce{[Ru(H2O)6]^3+}$, an octahedral complex. The electron configuration of the Ruthenium ion can either be $\mathrm{t_{2g}{}^5\,e_g{}^0}$ (low spin), or $\mathrm{t_{2g}{}^3\,e_g{}^2}$ (high spin), as shown in the pictures below:



You also need to know about CFSE (crystal field stabilization energy), which is the difference in the energy of the free metal ion, and the metal ion in the complex.

According to the electron configuration diagrams above, the CFSE for each configuration is:

$$\text{Octahedral }\mathrm{d^5}\text{ — low spin}\\
\text{CFSE} = -0.4\Delta_\mathrm{O} \times 5 + 2P = 2P - 2 \Delta_\mathrm{O}\\[1.5em]
\text{Octahedral }\mathrm{d^5}\text{ — high spin}\\
\text{CFSE} = -0.4\Delta_\mathrm{O} \times 3 + 0.6\Delta_\mathrm{O} \times 3 = 1.2\Delta_\mathrm{O} - 1.2\Delta_\mathrm{O} = 0$$

Where $P$ is the mean pairing energy, the energy required to overcome the repulsion between the electrons and pair them together. $\Delta_\mathrm{O}$ is the octahedral splitting parameter, which I assume you already know.

As you can see, a high spin $\mathrm{d^5}$ octahedral complex will not lead to a change in energy compared to a free ion. Therefore, the spin is determined by the low spin configuration. If the pairing energy is bigger than the octahedral splitting parameter, the metal ion will gain energy, and will rather be high spin. If the pairing energy is smaller than the octahedral splitting parameter, the metal ion will lose energy and will rather be low spin.

The octahedral splitting parameter is mostly affected by the ligand type (sigma, pi acceptor, pi donor). The pairing energy is mostly affected by the size of the d orbitals — the bigger they are, the smaller it is (because the electrons have more room, the repulsion between them is smaller).

4d and 5d transition metals have large d orbitals, so that the mean pairing energy is usually smaller than the splitting parameter, which results in low spin complexes regardless to the ligand type. 

Therefore, a $\ce{Ru^3+}$ ion, which is $\mathrm{4d^5}$, will form low spin octahedral complex, that has spin allowed transitions with intense colors.

About charge transfer. It's possible, but I personally don't know.