Let's take the simple system of a deuterium nucleus, that is the bound state of a proton and neutron.
It's tempting to think of the binding energy as something that has to be added to a proton and neutron to glue them together into a deuteron, and therefore that the deuteron must weigh more than the proton and neutron because it has had something extra added to it. However this is the exact opposite of what actually happens.
Suppose we start with a proton and a neutron at a large separation, and we let them go and allow them to attract each other. As they move towards each other their potential energy decreases, so because of energy conservation their kinetic energy increases. They accelerate towards each other in just the same way you accelerate towards the Earth if you jump out of a window. This means that when the proton and neutron meet they are moving very rapidly towards each other, and they just flash past each other and coast back out to a large separation. This is not a bound state. To form a deuteron you have to take energy out of the system so that when the proton and neutron meet they are stationary with respect to each other. Then they can bond to each other to form a deuteron.
And this is the key point. To form a bound state you need to take an amount of energy out that is equal to the kinetic energy gained as the two particles collide. This energy is the binding energy, which is 2.2 MeV for a deuteron. If you take the energy 2.2 MeV and convert it to a mass using Einstein's equation $E = mc^2$ you get a mass of $3.97 \times 10^{-30}$ kg, and if you compare the mass of a deuteron with the mass of a proton + neutron you find the deuteron is indeed $3.97 \times 10^{-30}$ kg lighter.
Just for completeness let's look at this the other way round. Suppose we start with a deuteron and we want to separate it into a proton and neutron. Because the two particles in a deuteron attract each other we need to do work to pull them apart, which means we need to add energy to the system. To pull them completely apart we need to add an amount of energy equal to the binding energy of 2.2 MeV. That means our separated proton and neutron have 2.2 MeV more energy than the deuteron, so the total mass of the separated particles has increased by 2.2 MeV and they are $3.97 \times 10^{-30}$ kg heavier than the deuteron.
Evidence that there are distinct protons and neutrons in nuclei starts with the Pauli term (pairing term) in the semiempirical mass formula of the liquid drop model.
Furthermore, all nuclei with even numbers of protons and neutrons have nuclear spin of zero. This is consisent with shells being filled with spin up and spin down pairs of nucleons, each pair resulting in net zero spin.
More generally, that experimental data are consistent with the Nuclear Shell Model is evidence that distinct protons and neutrons exist in the nucleus.
Also, the protons and neutrons are held together by exchange of pions. The exchange can result in the proton becoming a neutron and a neutron becoming a proton, so it is not that they exist entirely "as is".
See A reappraisal of the mechanism of pion exchange and its implications for the teaching of particle physics for furthur discussion of pion exchange.
Best Answer
This is basically a matter of energy scales. By analogy, you could ask why we don't take into account nuclear structure when we talk about chemistry. The answer is that the eV energy scale of chemistry is mismatched with the MeV energy scale of nuclear structure.
Nuclear matter has two phases. One is the phase we normally see, and the other is a quark-gluon plasma. The phase transition happens at a temperature on the order of 100 GeV per nucleon (at standard nuclear densities). Below the temperature of the phase transition, the quarks are strongly correlated, and those correlated groups behave in a way that's very similar to free neutrons and protons. To the extent that they don't quite have those properties, often we can subsume the discrepancies within adjustments to the parameters of the model. It's helpful in terms of practical computation that the fictitious neutrons and protons are nonrelativistic, which makes the theory much more tractable than QCD. If there are small relativistic effects, because the nucleons are moving at a few percent of $c$, these can also be subsumed within adjustments to the parameters.
By the way, it is actually possible to consider larger clusters to be the relevant degrees of freedom for nuclear structure. There is a model called the interacting boson approximation (IBA, also known as the interacting boson model, IBM), in which pairs of nucleons coupled to spin 0 or 1 are considered the degrees of freedom. It does pretty well in phenomonologically fitting the properties of many nuclei that are intractable in other models.
In a similar vein, there are alpha cluster models and ideas like explaining alpha decay in terms of preformation of an alpha particle, which then tunnels out through the Coulomb barrier. Pictures like these go back to the 1940's, and have considerable utility and explanatory power, although they can't really be microscopically correct, because they violate the Pauli exclusion principle.