No. E.g. $\psi=\begin{pmatrix}0&0&-1&1\\0&0&0&0\\1&0&0&0\\1&0&0&0\end{pmatrix}$ is nilpotent: $\psi^3=0$. If your $\phi=\begin{pmatrix}0&a&0&0\\-a&0&0&0\\0&0&0&b\\0&0&b&0\end{pmatrix}$ was $Q\psi Q^{-1}$, we would have $\phi^3=\begin{pmatrix}0&-a^3&0&0\\a^3&0&0&0\\0&0&0&b^3\\0&0&b^3&0\end{pmatrix}=0$, whence $a=b=\phi=\psi=0$, contradiction.
Edit (Re: your amended question below): If all (or even some) eigenvalues of $\psi$ are nonzero then yes, $\psi$ is conjugate to one of your $\phi$'s. It is equivalent but easier to classify adjoint orbits of the covering group $\mathrm{SL}(2,\mathbf C)$, whose Lie algebra consists of all
$$
Z=J(\mathbf z) := \frac1{2i}\begin{pmatrix}z_3&z_1-iz_2\\z_1+iz_2&-z_3\end{pmatrix},
\qquad \mathbf z = \mathbf a + \mathbf bi\in\mathbf C^3.
$$
The Lie algebra isomorphism maps $Z$ to $\psi=\begin{pmatrix}j(\mathbf a)&\mathbf b\\{}^t\mathbf b&0\end{pmatrix}$ where $j(\mathbf a)=\left(\begin{smallmatrix}0&-a_3&a_2\\a_3&0&-a_1\\-a_2&a_1&0\end{smallmatrix}\right)$.
Now $\det(\lambda\mathbf1-2iZ)=\lambda^2 - (\mathbf z,\mathbf z)=(\lambda+\zeta)(\lambda-\zeta)$ where $\zeta$ is a square root of $(\mathbf z,\mathbf z):=$ $z_1^2+z_2^2+z_3^2$ $=\|\mathbf a\|^2-\|\mathbf b\|^2+2i(\mathbf a,\mathbf b)$. So there are two cases:
if $(\mathbf z,\mathbf z)= 0$ then $Z$ is nilpotent, hence either 0 or (Jordan) conjugate to $\smash{\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)}$ $= J(i\mathbf e_1-\mathbf e_2)$ which maps to my nilpotent $\psi$ above (eigenvalues: 0).
if $(\mathbf z,\mathbf z)\ne 0$ then $Z$ is diagonalizable and hence conjugate to $\smash{\frac1{2i}\left(\begin{smallmatrix}\zeta&0\\0&-\zeta\end{smallmatrix}\right)}=J(\zeta\mathbf e_3)$ which maps to your $\phi$ if $\zeta = a + bi$ (eigenvalues: $\pm ai,\pm b$).
