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.
Best Answer
You can consider a thermocouple in its usual measurement role as a voltage source with a series resistance. The open-circuit voltage is a nonlinear function of two temperatures, the so-called cold junction(s) and the measuring junction. It can be expressed in terms of a nonlinear function with one parameter.
The series resistance is a function of the temperature along the wires but usually low enough it can be ignored. You can calculate the loop resistance by the usual formula R= \$\rho L \over A\$ for the two metals, or simply look it up in a table of thermocouple wire or extension leadwire from the material type, gauge and length. Extension leadwire may be made of a similar alloy to the thermocouple itself in the case of base-metal thermocouples such as J and K (Iron-Constantan and Chromel-Alumel) E and N, obviously that is a much less attractive proposition in the case of types R, S and B, and for different reasons, W, so the thermoelectric characteristics are approximated with different alloys (that are unsuitable for the measuring junction) for the extension leadwire.
The engineer designing the installation may choose a larger gauge of extension leadwire if the runs are long, in order to keep the resistance relatively low. If everything is within a few meters it usually doesn't matter.
The resistance is usually < 100 ohms, often closer to 10 ohms. Usually measuring and controlling instruments introduce a small current into the thermocouple to detect a broken sensor, maybe a few hundred nA. Say 200nA and a 20 ohm resistance will cause the sensor to read about 0.1°C high for a type K thermocouple. That offset will increase somewhat with temperature as the resistivity of the metals increases.
If the thermocouple junction is grounded (which is preferable in many situations) your measuring circuit should accommodate that using a galvanically isolated or differential front end.
In the old days, thermocouples were made with a defined resistance and fed a meter movement in the pyrometer (with some additional complexity we don't need to go into here), while that may still be true in a few cases, it's rare on industrial instruments. You should just assume the sensor has series resistance 0 < Rtc < 100\$\Omega\$ (in most cases).
To answer your questions: