I was in for a surprise. Knowing that there are no dipole (or, for that matter, monopole) gravitational waves, I was absolutely convinced that an axisymmetric system -- say, a pair of black holes in a head-on collision -- cannot possibly generate gravitational waves. Simple symmetry arguments, I thought, clearly show that generation of quadrupole (or higher multipole) radiation would require breaking the axial symmetry.
I was wrong, and there is literature to prove it.
To see why, I figured I should construct a simple example: two weights, connected by a spring, oscillating lengthwise without rotation. Simple enough, is it not.
Classical mechanics
The physics of this system in the context of classical mechanics is straightforward. Assuming that the two weights move around the $z$ axis and the spring is an ideal spring with relaxed length $L_0$ and spring constant $k$, the Lagrangian is
\begin{align}
L=\tfrac{1}{2}m\dot{z}_1^2+\tfrac{1}{2}m\dot{z}_2^2-\tfrac{1}{2}k(z_2-z_1-L_0)^2.\tag{1}
\end{align}
The corresponding Euler-Lagrange equations are, of course,
\begin{align}
\frac{\partial L}{\partial z_1}-\frac{d}{dt}\frac{\partial L}{\partial\dot{z}_1}=k(z_2-z_1-L_0)-m\ddot{z_1}&{}=0,\\
\frac{\partial L}{\partial z_2}-\frac{d}{dt}\frac{\partial L}{\partial\dot{z}_2}=k(z_1-z_2+L_0)-m\ddot{z_2}&{}=0.\tag{2}
\end{align}
A general solution is given in the form,
\begin{align}
z_{1,2}=\pm\tfrac{1}{2}L_0\pm\tfrac{1}{2}A\cos(\omega t+\phi),\tag{3}
\end{align}
i.e., symmetric oscillation with amplitude $A$, frequency $\omega=\sqrt{2k/m}$ and initial phase $\phi$ at $t=0$.
Gravitational radiation
To model gravitational radiation, we begin with Minkowski spacetime $\eta_{\mu\nu}$, slightly perturbed by $|h_{\mu\nu}|\ll 1$:
\begin{align}
g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}.\tag{4}
\end{align}
The standard route then is to define the trace-reversed perturbation $\bar{h}_{\mu\nu}=h_{\mu\nu}-\tfrac{1}{2}\eta_{\mu\nu}h$ and impose the Lorenz gauge condition $\partial^\mu\bar{h}_{\mu\nu}=0$, which then leads to the linearized Einstein field equations in the form of a sourced wave equation,
\begin{align}
\Box\bar{h}_{\mu\nu}=-\frac{16\pi G}{c^4}T_{\mu\nu},\tag{5}
\end{align}
with the conserved stress-energy tensor, $\partial_\mu T^{\mu\nu}=0$, acting as source. This leads to the retarded Green's function solution
\begin{align}
\bar{h}_{\mu\nu}(t,\mathbf{x})=\frac{4G}{c^4}\int d^3 x' \frac{T_{\mu\nu}(t-|\mathbf{x}-\mathbf{x}'|/c,\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|}.\tag{6}
\end{align}
We explore the far field, $|\mathbf{x}-\mathbf{x}'|\approx|\mathbf{x}|=R$, leading to
\begin{align}
\bar{h}_{\mu\nu}(t,\mathbf{x})\approx\frac{4G}{c^4R}\int d^3x' T_{\mu\nu}(t-R/c,\mathbf{x}').\tag{7}
\end{align}
In the slow motion limit, $v\ll c$, $T^{00}\sim \rho c^2$, $T^{0i}\sim \rho cv^i$, $T^{ij}\sim \rho v^i v^j$, defining the quadrupole moment $I_{ij}(t)=\int d^3x \rho(t,\mathbf{x})x_ix_j$, we have
\begin{align}
\frac{d^2 I_{ij}}{dt^2}=2\int d^3x T^{ij}+{\cal O}(v^2/c^2),\tag{8}
\end{align}
allowing us to trade the integral for $\ddot{I}_{ij}$. In the far zone, this gets us the leading-order waveform
\begin{align}
h_{ij}^\text{TT}(t,\mathbf{x})=\frac{2G}{c^4R}\ddot{Q}_{ij}^\text{TT}(t-R/c),\tag{9}
\end{align}
where $Q_{ij}=I_{ij}-\tfrac{1}{3}\delta_{ij}\text{tr}~I$ is the trace-free quadrupole.
The corresponding energy-flux, averaged over several wavelengths, yields a well-known formula:
\begin{align}
P=\frac{G}{5c^5}\langle\delta^{ik}\delta^{jl}\dddot{Q}_{ij}\dddot{Q}_{kl}\rangle.\tag{10}
\end{align}
The case of the two-mass system
The standard Newtonian symmetric trace-free (SFT) mass quadrupole moment for the two-mass system is
\begin{align}
Q_{zz}=\tfrac{1}{3}mr^2(t),\qquad Q_{xx}=Q_{yy}=-\tfrac{1}{6}mr^2(t),\tag{11}
\end{align}
with $z_{1,2}=\pm\tfrac{1}{2}r$, thus $r=L_0+A\cos(\omega t+\phi)$.
Plugging this in, time-averaging $(\langle\sin(\omega t+\phi)^2\rangle=1/2, \langle\sin(\omega t+\phi)\rangle=0)$ and using $L_0\gg A$, we get
\begin{align}
P=\frac{G}{15c^5}\,m^2L_0^2A^2\omega^6.\tag{12}
\end{align}
Given the internal energy of the oscillator, $E=\tfrac{1}{4}m\omega^2 A^2$, and
\begin{align}
P=-\dot{E}=-\tfrac{1}{2}m\omega^2 A\dot{A},\tag{13}
\end{align}
we can solve for $A$:
\begin{align}
A(t)=A(0)e^{-t/\tau},\qquad \tau=\frac{15c^5}{2GmL_0^2\omega^4}.\tag{14}
\end{align}
That is an exceedingly small time constant. (For a system with two 1-kilogram masses 1 meter apart oscillating at 1 Hz, it takes approximately $4\times 10^{32}$ times the present age of the universe for the amplitude to decrease by one $e$-fold.) But it is not zero. The axisymmetric system radiates.
Further calculation also reveals that the radiation is exclusively of the "+" polarization and its magnitude is proportional to $\sin^2\theta$, where $\theta$ is the polar angle.
In light of this, I am no longer shocked or surprised to learn that both a test particle falling into a black hole [1] and two black holes colliding [2] head on radiate. But the point is, no black holes, no strong fields, no event horizons are needed. Gravitational radiation, however weak, is present even in this simple case of a nonrelativistic oscillator. Axisymmetry constrains the form of the quadrupole tensor but does not force it to be constant in time.
References
[1] Davis, M. et al., Gravitational Radiation from a Particle Falling Radially into a Schwarzschild Black Hole, PRL 27(21) 1971
[2] Anninos, P. et al., Collision of Two Black Holes, PRL 71(18) 1993
