is there some kind of characterization of $h$ such that the domain $Q$ is convex
No, because the support function of $Q$ is equal to the support function of the convex hull of $Q$.
is there some kind of characterization of $h$ such that the domain $Q$ is strictly convex
Yes: if and only if $h$ is differentiable. In other words, if and only if the polar set of $Q$ has smooth boundary. It's worth noticing that the polar of polar set is the original set.
A corner in a convex set creates a line segment in the boundary of its polar, because the value of support function comes from the same point (the corner) for some interval of $\theta$. Conversely, suppose the boundary of a convex set contains a vertical segment to the right of origin. Then $h$ is not differentiable at $\theta=0$, because a small change of $\theta$ in either direction increases $h$ at a linear rate. (So, the graph of $h$ has a nonsmooth minimum like $|\theta|$.)
The general form of the above is known as duality of smoothness and rotundity, and it holds in all finite dimensional spaces.
Is there any characterization of $h$ such that the boundary $\partial Q$ admits nonvanishing curvature?
Yes: if and only if $h\in C^2$. I recommend Convex Bodies: The Brunn-Minkowski Theory by Rolf Schneider, specifically section 2.5 Higher regularity and curvature. For the two-dimensional case, see also page 2 of Lectures on Mean Curvature Flows by Xi-Ping Zhu.
Sketch. Introduce the Gauss map $\nu:\partial Q\to S^1$ which gives exterior unit normal vector at every boundary point. This map is defined when $Q$ has $C^1$ boundary. If $\partial Q$ is also strictly convex, then $\nu$ is injective. And if $\partial Q$ has nonvanishing curvature, then $\nu$ is a diffeomorphism -- indeed, this is if and only if, because the derivative of $\nu$ is the curvature of $\partial Q$.
It is convenient to define $h_Q$ as a function on $\mathbb R^2$, via $h_Q(x)=\sup_{q\in Q} x\cdot q$. Then one can check that $ h_Q(x)= x\cdot \nu^{-1}(x/|x|) $ and
$ \nabla h_Q(x)= \nu^{-1}(x/|x|)$.
Thus, the equivalence is: nonvanishing curvature $\iff$ $\nu $ is a diffeomorphism $\iff$ $h_Q\in C^2(\mathbb R^2\setminus \{0\})$.
Two conditions must be true for a convex function to be proper: (i) the epigraph must be non empty and (ii) never takes the value $-\infty$.
Since $f(x)$ is finite the epigraph is non empty.
If there is some value $y$ such that $f(y) = -\infty$ then it must be the case that $f(t) = -\infty$ for $t \in \operatorname{ri}(\operatorname{dom} f)$ (See Rockafellar, "Convex Analysis", Theorem 7.2. Since $f(x)$ is finite, we must have $f(t) > -\infty$ for all $t$.
To see that latter, suppose $f(y) = -\infty$. Since $x \in \operatorname{ri}(\operatorname{dom} f)$ there is some $z \in \operatorname{ri}(\operatorname{dom} f)$ and $\lambda \in (0,1)$ such that $x = \lambda y+(1-\lambda)z$ and hence $f(x) \le \lambda f(y) +(1-\lambda)f(x) = -\infty$ which is a contradiction.
Best Answer
After realizing the flaw in my previous, affirmative, answer: No such function exists.
If $f$ has a nonempty domain at every point, then it grows arbitrarily large, but since it is continuous, after sufficiently many iterations of log its range will always be $(-\infty,\infty)$. WLOG, take $f$ to be such a function (if any valid $f$ exists, all of its logarithms are valid too). It is strictly monotonically increasing (as it is convex and not bounded below), so we let $c_0<c_1<c_2<\ldots$ be the unique points at which $f$ attains the values $0,1,e,e^e,e^{e^e},\ldots$.
Now consider $g=\log(f)$. We have $g(c_1)=0, g(c_2)=1$, and that $g$ is convex and monotonically increasing. We also have $\lim_{x\to c_0}g(x)=-\infty$. But then consider the line from $(x,g(x))$ to $(c_2,1)$; as $x$ approaches $c_0$ from the positive direction, this line will drop below the point $(c_1,0)$, violating convexity.
It is possible for $f$ and all logs applied to it to be convex on all positive values, as outlined in my other answer, but not over its whole domain.