Suppose the Euler-Lagrange equation of a system $$\frac{\partial L}{\partial q}=\frac{d}{dt}\Bigg(\frac{\partial L}{\partial \dot{q}}\Bigg)$$ is known to be not invariant under the discrete transformation $t\to -t$. It's given. Also assume that nothing is known about the functional form of the Lagrangian $L(q,\dot{q},t)$.
Is it possible to infer from that information whether the Lagrangian (or action) will be invariant under $t\to -t$ or not?
Your question asks if T-asymmetry of the equations of motion implies T-asymmetry of the Lagrangian. That's equivalent to asking if T-symmetry of the Lagrangian implies T-symmetry of the equations of motion.
The answer is yes, by the principle of least action. Specifically, T-symmetry of the Lagrangian implies the paths $q(t)$ and $q(-t)$ have the same action. Then $q(t)$ is a stationary point of the action if and only if $q(-t)$ is, so if $q(t)$ solves the equation of motion, so does $q(-t)$.
To clarify questions in the comments, very explicitly, the facts are
I just came across this question and it is very interesting. If we re-question to be: If the equation of motion(EOM) is invariant under time reversal, is the Lagrangian as well? At first grant, these two bits seem to imply each other. Given $L(\dot x,x)=\dot x^2/2-V(x)$, the EOM is $\ddot x=-dV/dx$. Of course, under $t\rightarrow -t$, EOM and $L$ are invariant. Or, given $L(\dot x,x)=e^{\Gamma t}(\dot x^2/2-V(x))$, the EOM is $\ddot x+\Gamma \dot x+dV/dx=0$. Under $t\rightarrow -t$, EOM and $L$ are NOT invariant.
However, there is another class of Lagrangian called the multiplicative Lagrangian. We write the Lagrangian in the form $L(\dot x,x)=F(\dot x)G(x)$ with a constraint $m\ddot x=-dV/dx$. Solving for $F$ and $G$, we find
$L_{\lambda,k}(x, \dot{x}) = \left[k\dot x+m\lambda^2 \left( e^{-\frac{\dot{x}^2}{2\lambda^2}} + \frac{\dot{x}}{\lambda^2} \int_0^{\dot{x}} e^{-\frac{v^2}{2\lambda^2}} dv \right)\right] e^{-V(x)/m\lambda^2}$ ,
where $k$ has a momentum unit and $\lambda$ has a velocity unit. It is not difficult to show (but lengthy) that this Lagrangian $L_{\lambda,k}$ gives exactly $m\ddot x=-dV/dx$. Under the transformation $t\rightarrow -t$, EOM is invariant, but the Lagrangian $L_{\lambda,k}$ is not
$L_{\lambda,k}(x, -\dot{x}) = \left[-k\dot x+m\lambda^2 \left( e^{-\frac{\dot{x}^2}{2\lambda^2}} + \frac{\dot{x}}{\lambda^2} \int_0^{\dot{x}} e^{-\frac{v^2}{2\lambda^2}} dv \right)\right] e^{-V(x)/m\lambda^2}\equiv L_{\lambda,-k}$.
Note that the Lagrangian $L_{\lambda,-k}$ also gives the EOM: $m\ddot x=-dV/dx$.
We consider further the limit
$\lim_{\lambda\rightarrow \infty}L_{\lambda,k}=m\dot x^2/2-V(x)+k\dot x+m\lambda^2=L(\dot x,x)+k\dot x+m\lambda^2$. We recover the standard Lagrangian together with total derivative term and constant term. Of course, according to non-uniqueness of Lagrangian, these two terms do not affect the EOM. The point is that, commonly, we can decorate the Lagrangian BY HAND according to non-uniqueness. But in the context of multiplicative Lagrangian, these two terms come out naturally with the structure of the Lagrangian (by solving inverse problem of calculus of variations). Therefore, this particular Lagrangian could be an interesting case that If the equation of motion (EOM) is invariant under time reversal, the Lagrangian is not necessarily invariant.
