Your question is ultimately about scaling. Scaling is best explored by non-dimensionalizing equations.
Let's look at a free-fall equation of a mass $m$ in the absence of air friction :
$$
m \frac{d^2 z}{dt^2} = - m g
$$
Obviously the time to fall of a given height will depend on gravity $g$. But your question is, will it be the same up to a scaling factor in time? (Of course in this case you'll easily guess that it's the case -- but we'll go to more complex cases later)
You can non-dimensionalize it by defining a characteristic height $H$ and time $T$. Say $H$ will be the height of your camera's viewing field and $T$ the lag between two frames of your movie. You get (getting rid also of $m$):
$$
\frac{d^2 \tilde z}{d\tilde t^2} = - \frac{g T^2}{H}
$$
Here, $\frac{g T^2}{H}$ is a non-dimensional group that characterizes the ratio between inertia and gravity force. You can easily see that, if you double $T$ when $g$ is divided by 4, you get exactly the same number: the movie with $g/4$ with frames every $2T$ will thus look exactly the same as the one with $g$ and frames every $T$.
That's the same for a purely viscous flow, if there is no other force present at all. Rather than looking at the flow itself, let's look at a heavy particle sedimenting in the fluid: its equation is
$$
\frac{dz}{dt} = \frac{(\rho_p-\rho_f)gd^2}{18\eta}
$$
to which you can apply the same non-dimensional process.
Now let's introduce a second force, e.g. friction in the free-fall (or inertia in the above sedimenting particle). You have 3 terms in the equation, such as:
$$
m \frac{d^2 z}{dt^2} = - c \frac{d z}{d t} - m g
$$
This means that you'll build two non-dimensional groups: one will be $\frac{c T}{m }$ and the other $\frac{g T^2}{H}$. This time, if $g$ is changed, you cannot tune $T$ so as to keep everything unchanged as this would modify the other non-dimensional group. Only, if $c$ is so small that its effect is negligible for either $g$ or $g/4$ conditions, then the effect will not be visible in your movie and you can continue tuning $T$ according to $g$ only.
For a viscous flow, many forces in addition to viscous ones can come at play. Inertia is one, but if you look at a reasonnably-sized flow of honey or pitch, inertia will be very low in both cases.
What may actually be more relevant is surface forces (interfacial tension), which will not vary in direct proportion to viscosity. But you could try to find a sort of honey that would have the same viscosity to surface tension as pitch, and preserve the scaling!
