Formulation of the question
The only input we need from relativistic QFT is that it carries a unitary representation of the Poincaré group. For the purposes of this question, we can treat the atom as an elementary particle with spin $1$. The intrinsic spins of the electron and nucleus are ignored in this question, so the "intrinsic spin" of $1$ represents the orbital angular momentum of the bound electron.
Let $|m,\mathbf{p}\rangle$ denote a state in which the atom has momentum $\mathbf{p}$ and in which the eigenvalue of $L_z$ is $m$:
$$
\newcommand{\la}{\langle}
\newcommand{\ra}{\rangle}
\newcommand{\bfp}{\mathbf{p}}
\newcommand{\bfx}{\mathbf{x}}
\newcommand{\bfy}{\mathbf{y}}
\newcommand{\pl}{\partial}
\newcommand{\bfzero}{\mathbf{0}}
L_z|m,\bfp\ra = m|m,\bfp\ra.
\tag{1}
$$
Since the atom has spin $1$, the allowed values of $m$ are $m\in\{1,0,-1\}$. The question says that Alice prepares a state with $m=0$. I'll assume the atom is "at rest" in Alice's lab, but that's ambiguous in quantum theory, because the atom can't be localized in any finite region of space unless the state is a superposition of different momentum eigenstates. Using such a superposition would complicate the analysis, so I'll assume that Alice prepares the atom in the state that is arbitrarily close to a momentum eigenstate with eigenvalue zero ($\bfp=\bfzero$). I'll write this state as
$$
|\psi\ra = |0,\bfzero\ra,
\tag{2}
$$
with the understanding that this is really a superposition of different momenta (to make it normalizable) but concentrated in an arbitrarily small neighborhood of $\bfp=\bfzero$ (to simplify the analysis).
The goal is to determine the expectation value of a boosted version of $L_x$ in the state (2), which is the same as the expectation value of the original $L_x$ in a boosted version of the state.
The Lie algebra of the Lorentz group
The analysis becomes more intuitive after we modify the notation slightly. $L_z$ is the generator of rotations in the $x$-$y$ plane: it rotates the $x$ and $y$ directions and leaves the $z$ direction unchanged. For that reason, it is more natural to write $L_{xy}$ instead of $L_z$. Similarly, the generator of boosts in the $z$-direction is naturally denoted $L_{zt}$, because a boost is just a hyperbolic "rotation" in the $z$-$t$ plane, where $t$ is the time direction. The generators are antisymmetric with respect to exchanging the two indices, and since we want a unitary representation, they should also be hermitian as operators on the Hilbert space:
$$
L_{ab}^\dagger=L_{ab}
\hspace{2cm}
L_{ba}=-L_{ab}.
\tag{3}
$$
Now the Lie algebra of the Lorentz group can be written in a simple way:
\begin{align*}
[L_{ab},L_{bc}] &\propto L_{ac} && \text{if }a\neq c \\
[L_{ab},L_{cd}] &= 0 && \text{if $a,b,c,d$ are all distinct}.
\tag{4}
\end{align*}
In particular,
$$
[L_{xy},L_{zt}] = 0,
\tag{5}
$$
which says that $L_z$ (in the original notation) commutes with a boost in the $z$-direction.
Analysis and result
Let $U$ denote a unitary operator that implements a boost in the $z$-direction. Equation (5) says that this operator commutes with $L_z$, so we have
$$
L_z U|\psi\ra = UL_z|\psi\ra = 0,
$$
which implies
$$
U|\psi\ra \propto |0,\bfp\ra
\tag{6}
$$
for some momentum $\bfp$ in the $z$-direction. The goal is to calculate the expectation value of $L_x$ in this boosted state, which is equivalent to calculating the expectation value of a boosted version of $L_x$ in the original state.
Now, let $R_y$ denote a unitary operator that implements a reflection along the $y$-direction. To see what this reflection does to the state (6), remember that $L_z$ is the generator of rotations in the $x$-$y$ plane: it rotates the $x$ and $y$ directions and leaves the $z$ direction invariant. A reflection along the $y$-axis reverses the direction of such a rotation, so we have
$$
L_z R_y =-R_y L_z.
\tag{7}
$$
Similarly,
$$
L_x R_y =-R_y L_x.
\tag{8}
$$
For any momentum $\bfp$ with zero $y$-component, equations (1) and (7) imply $R_y|m,\bfp\ra\propto |{-m},\bfp\ra$. In particular,
$$
R_y U|\psi\ra\propto U|\psi\ra.
\tag{9}
$$
Since $R_y$ is unitary, we have
\begin{align*}
\la\psi|U^\dagger L_x U|\psi\ra
&=
\la\psi|U^\dagger R_y^\dagger R_y L_x U|\psi\ra
\\
&=
-\la\psi|U^\dagger R_y^\dagger L_x R_y U|\psi\ra
\\
&=
-\la\psi|U^\dagger L_x U|\psi\ra,
\end{align*}
where equations (8) and (9) were used in the last two steps. This says that $\la\psi|U^\dagger L_x U|\psi\ra$ is equal to its own negative, so it must be zero.
