A very general discussion-Not specific to a system:
The internal energy, $U$, of a system is a function of state, which means that its value only depends on the thermodynamic variables ($P, V, T)$ for example, at a given state (this means for a given set of values of these variables).
Let us make this more concrete: Imagine the system is in a thermodynamic state where the thermodynamic variables have the values ($P_i, V_i, T_i$) ($i$ stands for initial). At these values of the thermodynamic variables the internal energy has a value:
Internal energy at the initial state $i$: $U(P_i,T_i,V_i)$.
You can think of a gas at pressure, volume and temperature condition ($P_i, V_i, T_i$). Now imagine you change the thermodynamic variables to these ones ($P_f, V_f, T_f$) ($f$ stands for final). The internal energy now has a new value
Internal energy at the initial state $f$: $U(P_f,T_f,V_f)$.
In this process you have changed the internal energy of the system by an amount:
Change in U: $\Delta U= U(P_f,T_f,V_f)- U(P_i,T_i,V_i)$
I hope it is clear to observe that the system could have followed an infinitely large set of $(P,V,T)$-points, along an infinitely large number of different paths in order to go from state $i$ to state $f$. However, these are not, in any way, influencing by how much $U$ will change, you can take which ever path you please to go from state $i$ to state $f$. So the system has no memory of the intermediate states.
In mathematical terminology, this means that the differential change, $dU$, is a perfect differential and this is stated by the simple mathematical expression
$\oint_C dU=0$
It is very similar to the gravitational potential of the Earth, for example, which tells us that the amount of energy we need to spend to lift an object by 3m, does not depend whether we bring it straight vertically up or we follow some other path.
Your definition is a bit funky. I'll explain.
Explanation
As much as I like that you explicitly express internal energy $U$ as a function of state, for clarity I'm going to drop $\boldsymbol{R}$ for now.
We can't directly measure the absolute internal energy of a system, we can only infer its change by understanding some defined process. Consider the change in internal energy between two arbitrary states. By definition this is:
$\Delta U_{1 \rightarrow 2} = U_2 - U_1$
This applies equally to the case where we are considering internal energy relative to some reference state. We can express this as
$\Delta U = U - U_\text{ref}$
It doesn't matter what our reference state is, we can set $U_\text{ref} = 0$ so that
$\Delta U = U$
For a given state, internal energy is always relative to the internal energy of some other (reference) state. In practice, analysis of a non-trivial system will involve multiple states, we could choose any as the reference or pick an appropriate standard state; in any case setting the value to zero or dealing exclusively with $\Delta U$ has a similar effect.
For some arbitrary state of interest (rather than processes), I'd argue setting $U_\text{ref} = 0$ is not only convenient, it's semantically the most appropriate thing to do. We make it zero by definition rather than making it, say $12.7$ (which would mean $U = \Delta U + 12.7$). There's no such thing as the "true" reference state, only the one you define.
Visualisation
Here's an illustration to underline why the absolute value of the reference state is unimportant.
This is a generic visualisation relevant to two arbitrary states of interest.
The relative positioning of $\boldsymbol{R}_\text{ref}$ doesn't affect $\Delta U_{1 \rightarrow 2}$.
If we set $\Delta U_1 = 0$, we're defining $\boldsymbol{R}_\text{ref} \equiv \boldsymbol{R_1}$
$\boldsymbol{R}_\text{ref} \equiv \boldsymbol{R_1}$ implies $\Delta U_{1 \rightarrow 2} = \Delta U_2 = U_2$
Best Answer
While there are many variables that characterize a thermodynamic system, such as volume $V$, pressure $P$, particle number $N$, chemical potential $\mu$, temperature $T$ and entropy $S$, these are not all independent of each others! In fact, any thermodynamic potential (such as internal energy, free energy, enthalpy) can be written as functions of either three of these variables.
Thus, in the most general case, you will get something like $$U(T,P,N)$$ where you specify temperature, pressure, and number of particles.
I think it's easier to understand if you realize that pressure and volume are intimately linked, and then think about the effect of interactions: These should get stronger if you reduce the volume of the system so particles are closer together and thus (typically) have a higher interaction energy.
In an ideal gas, there are no interactions, so volume doesn't really have an effect on the internal energy: $$U = \frac{3}{2} N k T$$
But if you have interactions, they will give you a contribution that depends on volume and, thus, on pressure.
EDIT: As an example, the van-der-Waals equation describes a gas of weakly interacting particles. There, Wikipedia gives the internal energy as $$U = \frac{3}{2} N k T - \frac{a' N^2}{V}$$ where $a'$ is a parameter describing the interaction.