This is a great question, although unfortunately it turns out to be very difficult to interpret it in a way that allows a definite answer. The question is ambiguous because of the way mass is defined in relativity. From the way the question is posed, I assume the OP doesn't have a lot of technical background in relativity. However, there is no way to resolve the ambiguities in the question without getting pretty technical.
In relativity, "mass" really means mass-energy. Mass isn't additive. For example, a photon has zero mass, but consider a system consisting of one photon moving to the right and another photon of equal energy moving to the left. This system has a nonzero mass. This follows from the definition of inertial mass in special relativity according to the equation $m^2=E^2-p^2$, in units with $c=1$.
In GR, the source of curvature isn't mass-energy density, it's the stress-energy tensor. Some of the components of the stress-energy tensor correspond to pressure rather than density of mass-energy $\rho$.
The total mass (i.e., mass-energy) of a system in GR is not always a well-defined thing. For an arbitrarily chosen spacetime, there is no way to define the total mass. There are definitions of mass that work (i.e., are conserved and scalar) in special cases, such as an asymptotically flat spacetime. For example, there's the ADM mass.
If we want to define the mass-energy density $\rho$ at a point, we can do that. It's one of the components of the stress-energy tensor. However, there are a couple of limitations here: (1) under a Lorentz boost, a $\rho=0$ can transform into a $\rho\ne0$; (2) a singularity isn't a point in space, it's more like a point removed from space, so we can't define $\rho$ at a singularity.
So for a singular spacetime, we can't define the mass-energy density at the singularity, and in a typical, general case, there is no way to define the total mass, either. You could have a spacetime with a family of observers defined, one at each nonsingular point in spacetime, such that every one of these observers detects $\rho=0$; however, other observers in different states of motion might measure $\rho=0$. This ambiguity only goes away if the whole stress-energy tensor vanishes, i.e., if it's a vacuum solution (not just an electrovac solution like the Reissner-Nordström metric).
It's probably possible to have a singularity such that, in the rest frame of the singularity, $\rho\rightarrow0$ as you approach the singularity, but the pressure blows up to infinity. (This isn't consistent with the equation of state of any known form of matter, and it violates various energy conditions.) However, the statement that $\rho\rightarrow0$ will be false in other frames.
It is definitely possible to take a bunch of massless ingredients such as photons, mix them together (so that the collection as a whole has nonzero mass), and then let them collapse gravitationally into a singularity. But then the ADM mass of the singularity won't be zero.
There are curvature singularities and conical singularities. For any curvature singularity, the energy stored in the gravitational field surrounding the singularity will probably show up as a nonzero ADM mass. A conical singularity might be the best bet for an affirmative answer to the question if you want a zero ADM mass as well as a zero stress-energy tensor everywhere. I don't know for sure whether a spacetime with these properties exists in 3+1 dimensions. I don't think conical singularities can form by gravitational collapse in our 3+1-dimensional universe.