If two resistors were connected in parallel and one was heated, would current flow due to johnson noise?
Johnson noise is just the statistical variation of the voltage across a resistor that is due to the random motion of the charges within it.

If you work out the numbers, the amount of power that this represents is very very tiny &mdash; even when measured over a 1 GHz bandwidth, we're just talking about 4 picowatts (-84 dBm). If you wanted to produce 3.3V from this, you'd only get about 1 picoamp, assuming you could find a diode whose leakage is significantly less than that.

Note that in the case of two resistors with the same value, the Johnson noise power will divide equally between them, with half dissipated in the source (hot) resistor and half dissipated in the load (cold) resistor. But also note that any real circuit will have nonzero values of parasitic inductance and capacitance, and these will serve to limit the bandwidth of the noise transfer, causing more of the power to stay in the source and less to be transferred to the load.