Is it OK to use resistors with lower Ohm ratings if I decrease the input voltage?
You cannot use a 3V source with either of the LEDs specified except at very low currents.
See the C503D LED data sheet here  On page 7, Fig1 see the voltage versus current curve. Typically you need 3V at 10 mA and 3.2V at 20 mA. At only 5 mA you need 2.9V.   Note that the graphs show "typical" values and that real world values will vary across a range on either side of the typical value. The data sheet MAY specify maximum and minimum values as well as typical values for some parameters. 

The PDV-P503 light dependant resistor data sheet is here
The dark resistance is > 1 MOhm and the illuminated resistance (10 lux) is 12 to 58 kOhm.  To turn off Q1 (daylight conditions) the base of Q1 needs to be below say 0.3 Volt. So the R1:R2 divider needs to produce &lt; 0.3V relative to V1+ in light. R2 needs to be sized for the highest illuminated resistance value of R1 (as a lower value will turn off more easily.
For V1 = Vsupply = Vs. 
Vbase = Vs x R1/(R1+R2)
So R2 = R1 x (Vs/Vb -1)  

eg For Vs = 3V. R1 = 58 k (worst case). Vb = 0.3V
R2 = 58k x (3/0.3 -1 ) = 522k ~= 560k or 470k.
This will turn Q1 on BUT the 3V supply is too low for either LED and a very low value of R3 is required.

For Vs = 5V. R1 = 58 k (worst case). Vb = 0.3V
R2 = 58k x (5V/0.3V -1 ) = 908k or about 1 MOhm!.
Higher assumed values of Vb for V_base_turn_off_max will lead to somewhat lower R1 values.

R3 original is sized for I_LED ~= 1 mA !
[ (9V - 3V_Vled)/5k6 ]

R3 is sized using:
R = V/I = available voltage across R3 / I_LED
R3 = (Vs - Vsat_Q1 - VLED)/I_LED
eg for Vs = 5V, VsatQ1 = 0.2V, VLED = 3.1V, ILED  = 10 mA.
R3 = (5V - 0.2 - 3.1)/0.010 = 1.7V/0.010A = 170 Ohms.   

The original values assumed a value of R1 at turn on (Vb~~=0.6V) of about 16K (!).