I'm not sure I understand the question completely, but I'll work off of the comment
Of course the faster you go, the faster you reach max-q. So you must be able to say "by going this fast, you will reach max-q by this altitude" surely.
and take it to suggest that there may be a way to include altitude and therefore density implicitly in an expression rather than have it appear explicitly.
What does implicit mean? If my acceleration from rest towards a finish line at a distance $d$ is $a$, then using $x = \frac{1}{2} a t^2$ I can say my time will be $\sqrt{2d/a}$ without ever having talked about my speed. It's in there because I used calculus to integrate acceleration to get velocity, and to integrate velocity to get position $x$ in order to get the $x = \frac{1}{2} a t^2$ equation, but once I have that, velocity is inside there (implicit) even though it's not visible.
In a practical situation, probably the answer is no, because each launch has a different trajectory - some start turning sooner so that their vertical climb rate is lower, whereas others stay nearly vertical for a longer time.
But if we stick to a vertical launch for simplicity, and stick to the simple definition of dynamic pressure $q$ of
$$q = \frac{1}{2} \rho v^2$$
what then?
Let's say I can use a simple formula for the drag force:
$$F_D = -\frac{1}{2} \rho C_D A v^2$$
where the coefficient of drag $C_D$ is a constant. The problem of course is that it is not constant and varies a lot with both velocity and density as you go supersonic, and that behavior, for a realistic rocket, does not come from a simple formula but instead from careful measurements or very complex numerical simulations on computers. That's another source of "no".
Here's an example from @RussellBorogove's answer:
Also, see this answer for more reading.
Also, for gravity (I almost forgot!):
$$a_{Grav} = -GM_E/(R_E+x)^2$$
where GM_E is the standard gravitational parameter for Earth and $R_E$ the Earth's radius, and of course $a_{Grav}$ is the acceleration due to gravity (since mass drops out later).
But what if we were launching on a crazy planet where max-q happened at fairly low velocity and we could treat drag with a simple formula?
The acceleration $a$ is then given by
$$a = \frac{F_{tot}}{m(t)} = \frac{F_{Thrust} + F_{Drag}}{m(t)} + a_{Grav}$$
and with a fixed flow of propellant $\frac{dm}{dt} = \dot{m}$ we can say
$$m(t) = m_0 - \dot{m}t$$
and
$$F_{Thrust}=\dot{m} v_{ex}$$
where $v_{ex}$ is the exhaust velocity. That gives
$$a = \frac{\dot{m} v_{ex} + F_{Drag}}{m_0 - \dot{m}t} + a_{Grav}$$
or
$$a = \frac{\dot{m} v_{ex} - \frac{1}{2} \rho C_D A v^2}{m_0 - \dot{m}t} - \frac{GM_E}{(R_E+x)^2}$$
I still need to know the density for the definition of $q$, and we can use a simple scale height approximation which will look a bit like your plots of pressure versus altitude. Assuming the atmosphere's temperature is constant (which it is not) we can say density and pressure are always proportional. Then
$$\rho(x) = \rho_0 \exp(-x/h_{scale})$$
Plugging that back into the equation for acceleration, you get
$$a(t) = \frac{\dot{m} v_{ex} - \frac{1}{2} \rho_0 \exp(-x(t)/h_{scale}) C_D A v^2(t)}{m_0 - \dot{m}t} - \frac{GM_E}{(R_E+x(t))^2}$$
That's an equation that has all three of these: height, velocity, and acceleration ($x$, $v$ and $a$) and so solving for $v(t)$ and $x(t)$ in order to get
$$q(t)=\frac{1}{2} \rho(t) v^2(t) = \frac{1}{2} \rho_0 \exp(-x(t)/h_{scale}) v^2(t) $$
and then solving for
$$\frac{dq}{dt}=0$$
in order to find the maximum is going to be quite the challenge!
I'm not aware of an analytical (simple equation) solution for $v(t)$ and $x(t)$ even in this highly simplified case, so I am going to go out on a limb and say "no" there isn't a way to get time to max-q without really solving the whole problem numerically in a computer, getting the altitude and velocity from the trajectory, then plugging the numerical results back into the equation for $q$ to get its maximum.
