Single-ended to differential conversion and ADC application
I have read that the ADC converts the difference between positive and
  negative signals, but their difference never reaches 5V.


Your single-ended to differential amplifier has a voltage gain of 1, so with 5Vpp input the differential output is 5Vpp. To get 5Vpp on each ADC input you need a voltage gain of 2, producing a differential output of 10Vpp. 


  I would like a single supply system (5 V DC). But my analog signal is
  0-5V. Does this mean that it is not feasible to implement such a
  circuit since Vref reaches ~4 V or I could connect 5V to Vref and V+
  of the amplifier?


A single 5V supply will work as the reference voltage if the amplifier has rail-to-rail output. However there will be some nonlinearity close to the supply rails and it may be difficult to get sufficient reference voltage stability. Therefore it might be better use a lower reference voltage and reduce the ADC input voltage range to match. 

Generally it is not necessary (or even desirable) to 'max out' the ADC input range. Using a 4.096V reference with a differential signal range of 5Vpp (rather than the theoretical maximum of 8.192Vpp) loses less than 1 bit of resolution. However if you must have the full range of codes then simply adjust the amplifier gain to produce it.