@Numrok 's analysis makes certain assumptions which are then mentioned in the last paragraph of the answer.
For simplicity assume that the speed of the cars is measured in m/s rather than km/hr.
In the head on collision with both cars of the same mass and speed the final kinetic energy will be zero if the cars interlock as a result of the collision.
So the loss of kinetic energy $(= 2 \times \frac 1 2 \; m\; 30^2)$, which is a measure of the damage done due to permanent deformation, is twice the initial kinetic energy of one of the cars travelling at 30 m/s $(=\frac 1 2 \; m\; 30^2)$.
If the cars do not interlock then the loss in kinetic energy will be less.
In the other case assuming that the handbrake is off in the stationary car and the cars interlock after the collision the cars move off together at 30 /s with a total kinetic energy of $2 \times \frac 1 2 \; m\; 30^2$ but the initial kinetic energy was $\frac 1 2 \; m\; 60^2 = 4 \times \frac 1 2 \; m\; 30^2$ so the loss of kinetic energy is $2 \times \frac 1 2 \; m\; 30^2$ which is the same as before.
However with the brake on and/or friction present the loss of kinetic energy with both cars finishing up stationary is $\frac 1 2 \; m\; 60^2 = 4 \times \frac 1 2 \; m\; 30^2$ which is more than the head on collision with both cars initially moving at $30$ m/s.
Given that the $2 \times 30$ m/s collision will probably lose less than the maximum $2 \times \frac 1 2 \; m\; 30^2$ amount of kinetic energy whereas the $0+ 60$ m/s collision will lose less than $4 \times \frac 1 2 \; m\; 30^2$ but more than $2 \times \frac 1 2 \; m\; 30^2$ amount of kinetic energy I would say that the collision with the $60$ m/s car hitting the stationary car will probably do more damage to the cars.
As pointed out by @RahulJA the "damage" done to the occupants of the cars is much more difficult to quantify as their acceleration is a factor which must be considered.