Do I need to terminate extremely short traces (< 0.5 inches)?
First, as you mention, the critical parameter is usually the rise and fall times of your edges. You can estimate the effective "frequency knee" of your signal by

\$ f = \dfrac{\alpha}{t_r} \$

where \$t_r\$ is the faster of your rise and fall times. The parameter \$\alpha\$ is a kind of fudge factor; it depends on whether you measured the rise time as a 10%-90% or a 20%-80% value, and some authors give numbers between 0.5 and  0.8, but to be safe you could just use 1.0.

As Jippie discusses in his answer, if the wavelength associated with this frequency is more than 10x the trace length, you generally don't need to worry about transmission line effects. 

And in fact this is just how most CMOS and TTL drivers are intended to operate---except for certain specific types, they don't really have the current drive capability to drive a termination resistor of 50 or 75 Ohms.

Another complication is that most CMOS and TTL devices won't have a spec for rise and fall time. You'll have to estimate it from the drive current capability and the load capacitance:

\$ t_r \approx \dfrac{(V_h-V_l)C}{I}\$ 

Where I is the short circuit output current for your driver and C is estimated from your track geometry and the input capacitance of the load.

If you are using ECL parts, be aware that even if you don't terminate the transmission line, they still need a pull-down resistor to properly bias the output transistor.