Are there any known chemical properties of tritium water that make it unusually different from protium water?
Yes, $\ce{T2O}$ has been prepared and is available in significant quantity.  When relatively pure, the energy released by the radioactive decay process is so intense that $\ce{T2O}$ will boil.  It must be transported in a shielded, cryogenic dewar.

A significant difference between compounds containing an element bonded to protium, deuterium or tritium is the strength of those 3 bonds.  The $\ce{X-H}$ bond will be the weakest and $\ce{X-T}$ bond the stongest.  We can compare these bond strengths by measuring the relative rates at which the $\ce{X-H}$, $\ce{X-D}$ and $\ce{X-T}$ bonds are broken in a given isotopically substituted compound. These rate differences reflect what is known as a primary kinetic isotope effect. In carbon systems (X=C) the maximum primary isotope effects are roughly as follows:
$$\frac{K_H}{K_D} \sim 6-7$$
$$\frac{K_H}{K_T} \sim 13-14$$
In other words, a $\ce{C-H}$ bond may break as much as 13 times faster than a $\ce{C-T}$ bond.

These same effects will also be seen with $\ce{O-H}$, $\ce{O-D}$ and $\ce{O-T}$ bonds.  The primary kinetic isotope effects will be slightly smaller here because the magnitude of the effect is mass dependent and oxygen has a larger mass than carbon.  Still, as you pointed out, the effects can be disastrous in biological systems.  Biological systems cannot survive if the rates for key reactions are slowed down by such large factors.

The blue color of protium water ($\ce{H2O}$) is due to red light absorption around 700 nm. The frequency of an absorption is given by the following equation

$$\nu_e=\frac{1}{2\pi}\sqrt{\frac{k}{\mu}}$$

Here $\ce{\mu}$ is the reduced mass of the system (e.g. the bond involved in the vibration that is producing the light absorption) and is given as

$$μ=\frac{m_1 \cdot m_2}{m_1 + m_2}$$

where $m_1$ and $m_2$ are the atomic masses located at both ends of the bond.  We see that the reduced mass for an $\ce{O-H}$ bond is 16/17, while it is 32/18 for an $\ce{O-D}$ bond. Since, as shown above, the vibrational frequency is inversely related to the reduced mass, we would expect the $\ce{O-H}$ vibration to occur at higher frequency (shorter wavelength) than the $\ce{O-D}$ vibration.  Indeed, while $\ce{H2O}$ absorbs around 700 nm, $\ce{D2O}$ absorbs at higher wavelength (~1000 nm) and is colorless (reference, see p. 82). Given the even larger reduced mass for $\ce{T2O}$ (48/19), it's absorption should be shifted even further out of the visible range, so it too should be colorless.