Reps of the conformal group are still reps of the Lorentz group, that's correct. The operator content of a CFT contains scalars, currents, tensors, spinors etc. However, just because you have a local operator $\phi(x)$ or $\psi_\alpha(x)$ doesn't mean you have bosonic (resp. fermionic) particles in the theory.
Let's recall what we mean by "particle". It's an excitation of the vacuum. In a free theory, for example, we write a bosonic field $\phi$ as
$$\phi(x) \sim \int e^{ipx} a^{\dagger}(p) + c.c.$$
so acting on the vacuum you get a linear superposition of modes $$|p\rangle = a^{\dagger}(p) |0\rangle.$$ By acting several times with $\phi$, we can construct multi-particle states $|p,q,\ldots\rangle.$
There is a more general way of developing the particle picture, and that's through the spectral representation,
http://en.wikipedia.org/wiki/K%C3%A4ll%C3%A9n%E2%80%93Lehmann_spectral_representation or sec. 7.1 in Peskin-Schroeder.
This means that you write the propagator as
$$ < \phi(x) \phi(0) > \; = \int_0^\infty \rho(\mu^2) \Delta(x,\mu^2)$$
where
$$\Delta(p,\mu^2) = \frac{1}{p^2 - \mu^2 + i0}$$
in $d=4$ dimensions. $\rho(\mu^2)$ is called the spectral density and encodes which states contribute to the two-point function. It can be calculated as (see Wiki or Peskin-Schroeder)
$$\rho(p^2) = \sum_{\text{states } s} \delta^d(p - p_s) |\langle s | \phi \rangle|^2$$
with the sum running over all states in the theory and $p_s^\mu$ is the 4-momentum of state $s$. If you have a free boson with mass $m$, you can derive that
$$\rho(p^2) \sim \delta^d(p^2 - m^2).$$
Peskin-Schroeder plot a typical graph of $\rho$ for an interacting theory in Fig. 7.2. There, you will have a bunch of other terms for $p^2 > m^2$ as well, when $\phi$ overlaps with heavier states.
Anyway, back to CFT. In a free CFT (d = 4), the two-point function is given by
$$<\phi(x)\phi(0)> = \frac{1}{x^2}$$
and we have the spectral representation
$$ \rho(p^2) \sim \delta^4(p^2)$$
which we interpret as there being a single massless excitation.
But a generic scalar operator in CFT has some anomalous dimension $\delta$, so the two-point function will be
$$\frac{1}{x^{2+2\delta}}, \quad \delta > 0.$$
In that case, you get
$$ \rho(p^2) \sim (p^2)^{\delta-1}$$
which is continuous. You asked if "we [could] understand it as some sort of integral over masses that removes the scales of the theory?" - and you see that it's indeed what happens, in a quantitative way. There is no scale, only a parameter $\delta$ controlling the power-law scaling of the contribution of each state. As a consequence, there is no discrete set of states with a well-defined energy: $\phi$ just creates a big, scale-invariant blob of states.
That's essentially why people make a difference between "fields" and "operators". Superficially, both look the same: they are reps of the Lorentz group and generate the Hilbert space of the theory. However, in a normal QFT, fields give you particles and you can do scattering. Generically, in a CFT the operator don't have particle-like excitations, so we need to use a different term - "operator". A field in QFT is of course a local operator, but a local operator in CFT isn't necessarily a field.
