Would 3 op-amps be necessary?
Not in theory. In theory, you could just connect one end of the thermocouple to ground and then just feed the other end to a non-inverting amplifier. The problem, though, is noise pickup. Thermocouples have long wires, and those long wires act as antennas, picking up all sorts of junk. In most circuits, this wouldn't be a problem, but because thermocouples have such low voltages, the junk can easily overwhelm your actual temperature signal. By building an instrumentation amplifier, with 3 op-amps, you can remove (most of) this noise.
You may be able to get away with a single op-amp differential amplifier, but the large resistor values you'd need to use to get good input impedance would create a large amount of Johnson noise, which would wind up in your signal.
If you don't want to go for a proper amplifier, you'd need to use three op-amps. However, the matched resistors you'd need, plus the op-amps, may end up costing more than an instrumentation amplifier that uses, say, a single gain-setting resistor.
Also, have you thought about your cold-junction compensation? One of the issues with thermocouples is that they measure differential temperature; e.g. you have one junction at temperature A and another at temperature B, the thermocouple voltage is
(some constant K) * (A - B).
If you want to find out the absolute temperature of A, you need to know the temperature B.
Now, from your requirements you may actually be able to get away with a cheap hack. You can just assume that B is, say, 25C (roughly room temperature) and as long as B doesn't go outside the range 12.5C-37.5C, the temperature you get for A will be within 25C of A's actual temperature. You have enough error tolerance that I'd consider this viable.
If, though, the ambient temperature your circuit must operate in can go outside that temperature range, you will need to incorporate cold junction compensation. This consists, basically, of generating a voltage with the same temperature coefficient as your thermocouple, but relative to absolute temperature; in other words, you have
((some constant K) * (A - B)) + C. C would be equal to your constant K times B; as such it cancels out B and you end up just with
((some constant K) * (A - B)) + K*B = K*A - K*B + K*B = K*A.
The typical method for generating this voltage is via a diode. This is best done on an IC, and as such you may find that a thermocouple amplifier with a built-in cold junction compensator will do you much better than an op-amp, and in fact may cost less.