Given a particle with mass $m$ and four momentum $p=\left(p_x,p_y,p_z,\sqrt{|\vec{p}|^2+m^2}\right)$. Is it possible to reach every point in phase space, i.e. every combination of values $p_x,p_y,p_z\in\mathbb{R}$, by a simple Lorentz-Boost?
I'm guessing no, but don't know why.
Two boosts in orthogonal directions give you a boost and a rotation, so you are free to include rotations. But actually the answer is even simpler. Since the particle is massive, start in its rest frame where $p = (0,0,0,m)$ (note that we usually put time first, but I'm using your notation). Obviously you can boost this in any direction, so by rotational symmetry we can just consider the boosts along $x$. And it should be easy to convince yourself that boosts along $x$ can give you a momentum $p = (p_x,0,0,\sqrt{p_x^2 + m^2})$ for any $p_x \in \mathbb{R}$. So the answer to your question is yes.
Note also, be careful how you use words like "phase space". Since by definition, phase space is the space of states available, the answer to your question is trivially yes. The more interesting question is whether the phase space is all of $\mathbb{R}^3$, or just a subset.
Let $\vec{p}$ be the original spatial part of the four-momentum, and let $\vec{p}'$ be the desired spatial part of the four-momentum. Let us assume that for any $\vec{p}'$, there exists a boost vector $\vec{\beta}$ under which $\vec{p}$ maps to $\vec{p}'$. This will lead to a contradiction.
We can choose our coordinate system such that $\vec{\beta}$ points in the $x$-direction. We can then rotate our coordinate system about this $x$-axis such that $\vec{p}$ points in the $xz$-plane (i.e., $p_y = 0$.) Under the action of the boost $\vec{\beta}$, we must have $p'_y = 0$ as well. (Remember, we defined our coordinates such that $\vec{\beta}$ was in the $x$-direction.) But we assumed that $\vec{p}'$ could be any vector in $\mathbb{R}^3$; we therefore have a contradiction.
We conclude that a single pure boost cannot map an arbitrary spatial momentum $\vec{p}$ to another arbitrary spatial momentum $\vec{p}'$.
