Do hydrogen atoms not bound to carbon also appear in 1H NMR spectra?
Yes, with a but.

First the yes: they do appear, albeit at slightly different chemical shifts as, of course, the environment around a proton in $\ce{-CH2\!\;\!-}$ is slightly different to the electronic environment around the proton in $\ce{-NH2}$. The $\ce{N-H}$, $\ce{O-H}$, and $\ce{C-H}$ bonds all have slightly different electronegativies.

Now for the but: the thing that more significantly distinguishes $\ce{N-H}$ and $\ce{O-H}$ from $\ce{C-H}$ is their lability in protic solvents. Exchange of labile protons in the sample molecule with deuterons (i.e. deuterium ions, $\ce{^2H+}$) in the solvent can can remove the proton peak from the spectrum (this exchange is an equilibrium process, but as the solvent is in large excess the equilibrium lies over towards full deuteration of the sample's labile groups. This can actually be useful in identifying peaks: first run a sample in, for example, $\ce{CDCl3}$, then add a couple of drops of $\ce{D2O}$ to your NMR tube, give it a shake, and re-run the spectrum -- your labile protons will have vanished.

Edit 1: Furthermore, as @user1573870 points out in the comments, exchange of protons (exhanging protons for protons, rather than exchanging protons for deuterons),  on "the NMR timescale" can lead to broadening, and the resulting peak can be, while technically present, very difficult to spot. I'd suggest anyone doing NMR do background reading on that, because it's both useful and interesting (and I need to do some, because I'm out of practice with NMR).

Edit 2:To directly answer @Chewie's question whether "we have to count or omit them in the integration of the spectrum", you only count each of the peaks that are present. Using the example of ethanol, as asked in the comments, $\ce{CH3CH2OH}$ should give three peaks:


The $\ce{CH3}-$ peak should integrate to show three protons,
The $\ce{-CH2}-$ peak should integrate to show two protons, and
The $\ce{-OH}$ peak should integrate to show one proton.


If due to exchange with the deuterated solvent the $\ce{-OH}$ peak is not present, or if the $\ce{-OH}$ peak is 'there' but so broad as to be invisible to you (or so broad that it is difficult to integrate accurately), then you would only give the integrals for the other two peaks, and those two peaks should give integrals with a ratio equivalent to their respective number of protons, i.e. 3:2. If the $\ce{-OH}$ peak is not there, its proton does not just get thrown in with one of the other peaks/integrals (unless, sneaky proton, it just happened to have the same integral as another peak, and was hidden behind it, but for something as simple as $\ce{CH3CH2OH}$ that is unlikely).

The punchiest way to say all of thise would be that one does not "integrate spectra"; one integrates peaks.