I'm not sure I fully understand what you want, but here is at least a partial answer. You have $$ \det\left(\frac{d}{dt} \rho(t)\right) = (-i)^n \det(e^{-iHt}[H, \rho(0)]e^{iHt}) = (-i)^n \det([H, \rho(0)]), $$ so that the determinant of the derivative is a nonzero constant. Thus $\frac{d}{dt} \rho(t)$ is uniformly bounded away from the zero matrix.
A simple calculation gives $\frac{d^2}{dt^2}\rho(t)=(-i)^2e^{-iHt}[H,[H,\rho(0)]]e^{iHt}$. Since $\frac{d}{dt} \rho(t)$ is bounded away from zero and $\frac{d^2}{dt^2} \rho(t)$ is bounded, $\rho(t)$ cannot have a limit. In fact, you can establish a lower limit on its "oscillation" (assuming this is the kind of oscillation you want).
The oscillation estimate: Suppose $f:\mathbb R\to\mathbb R^N$ is a $C^2$ function so that $|f'|\geq a>0$ and $|f''|\leq A<\infty$ for some constants $a$ and $A$. (This applies in particular to your $\rho$.) By the fundamental theorem of calculus $$ f(T+\tau)-f(T) = \tau f'(T)+\int_0^\tau\int_0^sf''(r)drds. $$ Thus for $\tau>0$ $$ |f(T+\tau)-f(T)| \geq \tau|f'(T)|-\int_0^\tau\int_0^s|f''(r)|drds \geq a\tau-\frac12A\tau^2. $$ Therefore $|f(T+\tau)-f(T)|$ has a bound from below uniformly in $T$ if $\tau<2a/A$. I could take this behavior as a definition of endless oscillation.