You can interpret that Bode plot as the response to a sinusoidal input even for unstable systems, if instead of "steady state" you say "with initial conditions exactly zero".
For a stable system, those two interpretations coincide. Remember that the space of all solutions of an inhomogenous system of linear differential equations $f'(t) = Af(t) + g(t)$ can always be written as $ \lambda_1f_1(t) + \ldots + \lambda_nf_n(t) + f_p(t) $ where $f_1,\ldots,f_n$ are (linearly independent) solutions of the homogenous system $f'(t) = Af(t)$, and $f_p$ is one particular solution of the inhomogenous system. Also remember that $\lambda_1,\ldots,\lambda_n$ are determined by the initial condition of the system. Now, if the system is stable, all the $f_1,\ldots,f_n$ decay exponentially. Thus, no matter what the initial conditions are, if $t$ gets large enough, the solution will basically just be $f_p(t)$.
For an unstable system, some of the $f_1,\ldots,f_n$ grow exponentially (or are constant), so you cannot count on their influence to vanish at some point. You can, however, simply decide to set $\lambda_1=\ldots=\lambda_n=0$ in the solution, and hence look only at $f_p$.
Take, for example, the simple system $ f'(t) = f(t) + e^{i\omega t} $ The corresponding homogenous system has the solution $f_1(t)=e^t$ and is thus unstable. A particular solution of the inhomogenous system is $ f_p(t)=-\frac{1+i\omega}{1+\omega^2}e^{i\omega t} $ and the frequency response (i.e. what you plot in a Bode plot) is therefore $ A(\omega) = -\frac{1+i\omega}{1+\omega^2} $
Note how it is perfectly reasonable to interpret this as the attenuation of the sinusoidal input $e^{i\omega t}$ - it is, after all, derived from an actual solution of the differential equation. You just won't be able to measure it, because in every actual experiment the initial condition (i.e. $\lambda_1$) will never be exactly zero - and once it differs from zero only the slightest, $\lambda_1f(t)=\lambda_1e^t$ will quickly dominate the result.