Do you intend to sell this device? If so, prevention is guaranteed by the fact that you'll never get FCC approval for it.
If you're approaching it from a security standpoint, imagining a target which is relying on NFC restrictions of 1 meter vs. several meters to protect valuable data, you could create such a device. It would be expensive, and thoroughly illegal. If your target has data that's worth more than the cost to build such a device, then they're probably not going to keep it in a format where it could be accessed via NFC.
A recommended approach would be to use an optocoupler followed by a comparator (eg. LM339), or better, an integrated part such as the Fairchild Semi FODM8071 logic gate output optocoupler.
The reason the optocoupler is recommended:
There is likely to be a ground potential difference over a 50 meter cable, also the possibility of picking up EMI over the long cable. The optocoupler eliminates any ground loop / potential mismatch concerns, as well as any need to precisely match the sensor's supply voltage to the microcontroller's.
The use of the opto will allow a higher voltage to be used for the sensor circuit, reducing EMI noise sensitivity.
An added benefit of the specific Fairchild part suggested above is its high noise immunity. This will result in a more stable signal acquisition, important given the distances involved.
FODM8071 is a 5-pin leaded SMT part, so using it is essentially like not having to build any additional circuit - you could wire the part and its few support discrete components up deadbug style, if you like, or put them together on a proto-board PCB.
Best Answer
I don't think so. Voyager 1 is a great example of this - it uses phase modulation (similar to frequency modulation) and it has been able to transmit messages to earth that are ridiculously long distances away.
As always, the "link loss" between transmitter and receiver is paramount in understanding how much attenuation occurs between transmitter and receiver. This is: -
link loss (dB) = = 32.45 + 20\$log_{10}\$(f) + 20\$log_{10}\$(d)
Where f is in MHz and d is in kilometres. This equation tells you how many dB of power loss you can expect at a given distance with a given carrier frequency.
What you may be influencing your question is that ionospheric bounce occurs quite well in the lower frequencies that are dominated by public AM transmission such as a local radio station. AM is used not because it gets further but because the thousands of radio receivers tuned-in have a simpler job demodulating the transmission i.e. it is cost driven - non-suppressed carrier AM is easy and cheap to decode, hence the extremely simple crystal set radios of the last century.