Yeh!, badly expressed, in the forward direction, the base is always between +.6 and + .8 of a volt with respect to the emitter on a NPN BJT. On a PNP, such as your Q1, its always negative. So the emitter is nailed to + 12V, so the base is .8 V negative i.e. 11.2 V, but your 555 output is even more negative at 9V, so there is a voltage drop of 2.2V driving current through the 2K2 resistor, giving an Ib of 1mA.
If you just change the resistor, say from the 2K2 to 220K, the Ib would still be 2.2/220 = 1/100 mA, which is better but the gain of Q1 might be 100, giving a Ib for the darlington of 1mA, X the 1000 gain of the darlington = 1 A, quite a lot for a transistor that is mean't to be off. And when the 555 output goes to earth the voltage drive to Q1 base is 11.2V, and the current is now 11.2/220 ~ .2MA, which is a bit low for the darlington to saturate.
What I would do is to check the output of the 555, to make sure that it is working and not broken, i.e. it goes to 80% of the +12V (or whatever) and not only 5V. Suppose its 8.2V. Now at this voltage you want to make sure Q1 is off, so its Vbe must be less then .6, say .4V. So if the base is at 8.2, the emitter must be less then 8.6V. So if you put a 100 ohms between the emitter and the +12V, its got to drop 12 - 8.6V = 3.4V, so the current through it must be 3.4/100 = 34 mA. So connect another resistor between the emitter and the 100 ohms, to earth, this resistor must drop 8.6 V @ 34 mA, so its value is 8.6/.034 = ~250 ohms. Or much better use enough diode in series with the base to make sure that the 3.4 V cannot turn them on, 4 X .8 = 3.2V too low, 5 X .8 V = 4V which is OK.
Frank