Consider your surface of revolution with the following parametrization, taking $z$ as the axis where the curve revolves,
$$\pmb{\mathrm{x}}(u,v)=(f(u)\cos v,f(u)\sin v,g(u)).$$
As you pointed out, for a such a parametrization we can assume that $f(u) > 0$, and also we can assume that the profile curve $u\mapsto(f(u), 0, g(u))$ is unit-speed, that is, $f'(u)^2 + g'(u)^2 = 1$.
Then,
$$\pmb{\mathrm{x}}_u=(f'(u)\cos v, f'(u) \sin v, g(u)),\;\pmb{\mathrm{x}}_v=(−f(u) \sin v, f(u) \cos v, 0).$$
Hence,
\begin{align}
&E=\|\pmb{\mathrm{x}}_u\|^2= f'(u)^2 + g'(u)^2 = 1,\\
&F=\langle\pmb{\mathrm{x}}_u,\pmb{\mathrm{x}}_v\rangle= 0,\\
&G =\|\pmb{\mathrm{x}}_v\|^2 = f(u)^2.
\end{align}
So the first fundamental form is given by
$$du^2 + f(u)^2dv^2.$$
We also have that
\begin{align}
\pmb{\mathrm{x}}_u\times\pmb{\mathrm{x}}_v&= (−f(u)g'(u) \cos v,−f(u)g'(u) \sin v, f(u)f'(u)),\\
\|\pmb{\mathrm{x}}_u\times\pmb{\mathrm{x}}_v\|&= f(u),\\
\pmb{N}&= {\pmb{\mathrm{x}}_u\times\pmb{\mathrm{x}}_v\over\|\pmb{\mathrm{x}}_u\times\pmb{\mathrm{x}}_v\|} = (−g'(u) \cos v,−g'(u) \sin v, f'(u)),\\
\pmb{\mathrm{x}}_{uu}&= (f''(u) \cos v, f''(u) \sin v, g''(u)),\\
\pmb{\mathrm{x}}_{uv}&= (−f'(u) \sin v, f'(u) \cos v, 0),\\
\pmb{\mathrm{x}}_{vv}&= (−f'(u) \cos v,−f(u) \sin v, 0),\\
L&= \langle\pmb{\mathrm{x}}_{uu},\pmb{N}\rangle = f'(u)g''(u) − f''(u)g'(u),\\
M&= \langle\pmb{\mathrm{x}}_{uv},\pmb{N}\rangle = 0,\\
N& = \langle\pmb{\mathrm{x}}_{vv},\pmb{N}\rangle = f(u)g'(u),
\end{align}
so the second fundamental form is
$$(f'(u)g''(u) − f''(u)g'(u))\;du^2 + f(u)g'(u)\;dv^2.$$
Now, we use the well-known expression for $H$ given by
$$H={1\over2}{EN-2FM+GL\over EG-F^2},$$
so we get
$$H={1\over2}\left(f'(u)g''(u)-f''(u)g'(u)+{g'(u)\over f(u)}\right).$$
So far, so good.
Suppose now that for some value of $u$, say $u = u_0$, we have $g'(u_0)\neq 0$. Since $g'(u)$ is continuous, we have that $g'(u) \neq 0$ for $u$ in some open interval containing $u_0$. Let $(\alpha, \beta)$ be the largest such interval. Supposing also that $u\in(\alpha, \beta)$, the condition $f'(u)^2 + g'(u)^2 = 1$ gives (by differentiating with respect to $u$) $f'(u)f''(u)+g'(u)g''(u)=0$, and then
\begin{align}
(f'(u)g''(u)-g'(u)f''(u))g'(u)&=\\
&=-f'(u)^2f''(u)-f''(u)g'(u)^2\\
&=-f''(u)(f'(u)^2+g'(u)^2)=-f''(u),
\end{align}
so $f'(u)g''(u)-g'(u)f''(u)={-f''(u)\over g'(u)}$ and we get
$$H={1\over 2}\left({g'(u)\over f(u)}-{f''(u)\over g'(u)}\right).$$
Again by $g'(u)^2=1-f'(u)^2$, the surface is minimal if and only if
$$f(u)f''(u)=1-f'(u)^2.$$
This is an ODE that can be solved by taking $h=f'$ (I will stop writing the dependance on $u$) and noticing that
$$f''(u)={dh\over dt}={dh\over df}{df\over dt}=h{dh\over df}.$$
Hence, the ODE is now
$$fh{dh\over df}=1-h^2.$$
We assumed that $g'(u)\neq 0$, so $h^2\neq 1$ (as $h^2+g'(u)^2=1$). This allows us to rearrange and integrate the equation
$$\int {h\over 1-h^2}\;dh\;=\;\int {1\over f}\;df,$$
which gives us
$${1\over \sqrt{1-h^2}}\;=\;af\;,$$
(where $a>0$ is a constant) and then
$$h\;=\;{\sqrt{a^2f^2-1}\over af}.$$
With this, since $h=f'$, integrating again
$$\int {af\over\sqrt{a^2f^2-1}}\;df\;=\;\int\;du.$$
If we solve this integral equation for $f$ we obtain
$$f={1\over a}\sqrt{1+a^2(u+b)^2},$$
where $b$ is a constant that we can assume that is zero by taking the change of parameter $u\mapsto u+b$, so we have
$$f={1\over a}\sqrt{1+a^2u^2}.$$
With this expression for $f$, we can compute $g$ as follows
\begin{align}
&g'^2=1-f'^2=1-h^2={1\over a^2f^2}\;,\\
&g'=\pm{1\over\sqrt{1+a^2u^2}}\;,\\
&g\;=\;\pm{1\over a}\sinh^{-1}(au)+c,
\end{align}
for some constant $c$.
We will consider the term $au$ so we can write $f$ in terms of $g$:
$$au=\pm\sinh(a(g-c))\;,$$
so
$$f={1\over a}\cosh(a(g-c)).$$
By this, the profile curve of the surface is
$$x={1\over a}\cosh(a(z-c))$$
where, by a translation along the revolving axis, we can assume that $c=0$, obtaining the equation of a catenary.
So far, we have shown that the open subset of our surface corresponding to $u\in(\alpha,\beta)$ is part of the catenoid, for in the proof we used in an essential way that $g'(u)\neq 0$. This is why the proof has so far excluded the possibility that we are in the case of a plane.
Now, suppose that $\beta < \infty$. Then, if the curve is defined for $u\geq\beta$, we would get $g'(\beta)=0$. Otherwise $g'(u)$ would be non-zero on an open interval containing $\beta$, which would contradict the assumption that $(\alpha,\beta)$ is the largest open interval containing $u_0$ on which $g'(u)\neq 0$. Right, but the formulas above show that
$$g''^2={1\over a^2f(u)^2}={1\over 1+a^2u^2}$$
for $u\in(\alpha,\beta)$, and since $g'(u)$ is continuous, $g'(\beta)=\pm(1+a^2\beta^2)^{-{1\over 2}}\neq 0$.
This is a contradiction and then the profile curve can not be defined for $u\geq\beta$. In the case $\beta=\infty$, we obtain the same.
We should now proceed with a similar argument for $\alpha$, showing that $(\alpha,\beta)$ is the entire domain of definition of the profile curve. Hence, the whole surface of revolution is an open subset of a catenoid.
The remaining case is that in which $g'(u) = 0$. But then $g(u)$ is a constant, say $d$, and the surface would be an open subset of the plane $z = d$.
Best Answer
Perhaps an Euler spiral (also known as a clothoid)? Quoting the Wikipedia entry for "Euler spiral": An Euler spiral is a curve whose curvature changes linearly with its curve length.
In particular, in terms of your notation, reframe $\gamma$ as being the angle of turn after traversing length $t$. Then compare with this formulation (at the Wikipedia entry) which uses $\theta$ where you have used $\gamma$ and $s$ where you have used $t$. What you have called $(\gamma_x, \gamma_y)$ turn out to be obtainable by Fresnel integrals. Also see this question and answers from it.