Why does the wavefunction of a particle spread out after a measurement? Quantum mechanics states th

It is a mathematical phenomenon, it happens due to time-dependent Schroedinger equation. It is similar to diffusion equation, localized peaks in tend to dissipate according to both equations. So any localized wavepacket spreads unless that spreading is stopped by a potential rise.
This behaviour can be most easily shown and quantified for free particle. If particle is in some potential well, that well will slow down or even stop the spreading.

Assuming by "spread" you mean how "wide" the distribution of $\mid <!--\; \mid \; -->\mathrm{\psi <!--\; \psi \; -->}(x,t){\mid <!--\; \mid \; -->}^2$ (or for momentum space, the spread of $\mid <!--\; \mid \; -->\mathrm{\psi <!--\; \psi \; -->}(k,t){\mid <!--\; \mid \; -->}^2$).Quantum mechanics states that the wave packet of a particle "spreads-out" in position again after a measurement on this particle has been made.
Before and after measurement, the system evolves as described by the non-relativistic time-dependant Schrodinger equation
$i\frac{\mathrm{\partial <!--\; \partial \; -->}\mathrm{\psi <!--\; \psi \; -->}(x,t)}{\mathrm{\partial <!--\; \partial \; -->}t}=\stackrel{^<!--\; ^\; -->}{H}\mathrm{\psi <!--\; \psi \; -->}(x,t)$
The initial conditions for the system after the first measurement, are dependant on how you performed this measurement that caused the wave-function to collapse. So if you make the exact same measurement the same way, a very short time after the first, you will get a similar result to the first measurement. If you measured position, you will get sharply peaked distribution (Dirac delta function) in position space $\mathrm{\delta <!--\; \delta \; -->}(x-<!--\; -\; -->x^{\prime })$ centred at $x^{\prime }$ causing a "spread", or a large uncertainty in the momentum distribution. The wave packet does become wider.
"Is this spreading or "dispersion" caused by inherent properties of the wavefunction itself (eg according to the Uncertainty Principle), or is it due to the interaction with the environment (or both)?"
The wavefunction on its own has no physical properties. It is its square that has physical meaning. The uncertainty principle tells us that the more accurately we measure position, the less accurate we can know its momentum, and vice-versa. How it is measured (with what apparatus, how we do it etc) can affect the accuracy we get for either quantity, but this complementarity between accuracy of position/momentum is still described by the uncertainty relation. So the environment or apparatus is important only in terms of accuracy, which will obviously affect the amount of "spread".
"Does the wavefunction of a free, non-interacting particle also show this behaviour?"
Yes it does. The Heisenberg uncertainty relation still holds. And solving the Schrodinger equation will give you Gaussian distributions in both position and momentum space. Again, in free space, if you describe the particle with a wave packet, and you wanted to increase the accuracy in momentum, the position uncertainty does become wider.