Why is water's IR spectra indistinguishable from the IR spectra of a solution of glucose in water?
I see the spectra in the following way:

The sample and measurement: glucose in water and IR absorption measurement

Considering free glucose molecule, we should expect the vibrations from $C-H$, $O-H$, $C-O$ and $C-C$ bonds and other associated vibrations.
The approximate vibrational frequencies are :
a) $C-H$ = ~2800 - 3300 $cm^{-1}$ (stretching)
b) $O-H$ = ~3800 - 3200 $cm^{-1}$ (stretching)
c) $C-O$ = ~1200 - 1030 $cm^{-1}$ (stretching)
d) $C-C$ =  usually not good for interpretation
e) $C-H$, $CH_{2}$ and $O-H$ bend(in plane) = 1500-1200 $cm^{-1}$
and now considering water,
the $O-H$ stretching band  spectrally overlaps with the bands of glucose (~3750 - 3200). The combination band (~2300) and scissor mode (~1600) are also relatively broad and strong. Due to this spectral overlap and also larger infrared  absorption cross-section of water, glucose is not clearly identifiable in solution.

Instrument and spectral quality : For obtaining a high resolution spectra with good s/n ratio the instrument should be good (FT-IR) and long exposure is needed.
The spectra shown by @user1155386 has the $x-axis$ as nanometers (not wavenumbers  ! ), indicating that the spectral resolution is low. It is noisy showing that exposure time can be improved (the exposure time shown there is only 10 milliseconds).


However, still it would be difficult (but not impossible) to differentiate glucose solution and water. The main hurdle would be get a good spectra and the reason 1. mentioned above.
Additionally, I should mention Raman spectroscopy is better for aqueous solution measurements.