This answer is based on this article by A. Ungar.
Ungar computed the Thomas rotation formula which is almost what you need. I'll describe the general procedure, and in some cases, I'll refer you to Ungar for the proof. I'll express (just like Ungar), the Boosts in terms of velocities rather than momenta. If you wish you can repeat the exercise with the momentum parametrization. From Wikipedia we have
$$B(\mathbf{v}) = \begin{bmatrix}
\gamma_{\mathbf{v}}& \frac{\gamma_{\mathbf{v}}}{c}\mathbf{v}^t \\
\frac{\gamma_{\mathbf{v}}}{c}\mathbf{v} & 1_{(3\times3)}+
\frac{\gamma_{\mathbf{v}}^2}{1+\gamma_{\mathbf{v}}}
\frac{\mathbf{v}\mathbf{v}^t}{c^2}
\end{bmatrix}$$
The key point is finding the Wigner rotation is the observation that every Lorentz transformation can be decomposed (nonuniquely) as a product of a Boost and a rotation:
$$\Lambda = B(\mathbf{u}) R$$
Any choice of the decomposition method will do, but we need to work in a fixed method of decomposition. I'll describe to you a possible method at the end of the answer
Now, when we multiply two boosts, we get a boost with the relativistic addition velocity + a rotation (often referred to as the Thomas rotation):
$$ B(\mathbf{u}) B(\mathbf{v}) = B(\mathbf{u}\oplus \mathbf{v}) \mathrm{Tom}(\mathbf{u},\mathbf{v})$$
Where $\mathrm{Tom}$ is a rotation matrix
Ungar found the general solution for the Thomas rotation (Equation : (13) in the article)
$$\mathrm{Tom}(\mathbf{u},\mathbf{v}) = B(-\mathbf{u}\oplus \mathbf{v}) B(\mathbf{u}) B(\mathbf{v})$$
Now, we are in a position to solve the equation for the Wigner rotation. We need to solve:
$$ \Lambda B(\mathbf{v}) = B(\Lambda \mathbf{v}) W$$
for $W$. We parametrize $\Lambda$, we get for the left handside:
$$\begin{align*}
\Lambda B(\mathbf{v}) & = B(\mathbf{u}) R B(\mathbf{v}) \\ & = B(\mathbf{u}) R B(\mathbf{v}) R^{-1} R \\ & = B(\mathbf{u}) B(\mathbf{Rv}) R \\ & = B(\mathbf{u}\oplus R\mathbf{v}) \mathrm{Tom}(\mathbf{u},R\mathbf{v}) R
\end{align*} $$
and for the right hand side
$$ B(\Lambda \mathbf{v}) W = B(B(\mathbf{u}) R\mathbf{v}) W = B(\mathbf{u}\oplus R\mathbf{v}) W$$
Thus:
$$W = \mathrm{Tom}(\mathbf{u},R\mathbf{v}) R$$
What is left is to describe a specific parametrization of a general Lorentz matrix into a boost and a rotation:
We need to find $B(\mathbf{u})$ and $R$ such that:
$$\begin{bmatrix}
\lambda_0 &\mathbf{\xi}^t \\
\mathbf{\eta} & \Lambda_1
\end{bmatrix} = \begin{bmatrix}
\gamma_{\mathbf{v}}& \frac{\gamma_{\mathbf{v}}}{c}\mathbf{v}^t \\
\frac{\gamma_{\mathbf{v}}}{c}\mathbf{v} & 1_{(3\times3)}+
\frac{\gamma_{\mathbf{v}}^2}{1+\gamma_{\mathbf{v}}}
\frac{\mathbf{v}\mathbf{v}^t}{c^2}
\end{bmatrix} \begin{bmatrix}
1&0 \\
0 &R_1
\end{bmatrix}$$
We observe that $R_1$ needs to satisfy the following relation
$$\mathbf{\xi} = R_1 \mathbf{\eta}$$
Thus $R_1$ needs to rotate the 3-vector $\mathbf{\mathbf{\eta}}$ into the 3-vector $\mathbf{\xi}$
The solution (using a similar method as Ungar) can be written as:
$$R_1 = 1_{(3\times3)}+ \sin(\theta) \Omega + (\cos(\theta-1)) \Omega^2$$
Where $\theta$ is the angle between the vectors $\mathbf{\eta}$ and $\mathbf{\xi}$
$$\sin(\theta) = \frac{\mathbf{\eta} \times \mathbf{\xi}}{|\mathbf{\eta}||\mathbf{\mathbf{\xi}}|}$$
and
$$\Omega_{ij} =\frac{ \eta_i \xi_j - \xi_i \eta_j}{|\mathbf{\eta}||\mathbf{\xi}|}$$
Best Answer
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.