For any real-valued function $h$, $\alpha\in[0,1]$ and $x,y$ in the (convex) domain, let
$$
D(h,\alpha,x,y)=\alpha h(x)+(1-\alpha)h(y)-h[\alpha x+(1-\alpha)y].
$$
Convexity for $h$ means $D(h,\alpha,x,y)\geq 0$ for all $\alpha,x,y$.
For your situation, one sufficient condition for $f-g$ to be convex is that
$$
D(f,\alpha,x,y)\geq D(g,\alpha,x,y)\tag{i}
$$
for all $\alpha,x,y$. (I think of this as $f$ being "more convex" then $g$.) Note that (i) doesn't impose convexity directly on $f$ and $g$. For example, $f(x)=-x^2$ and $g(x)=-2x^2$ satisfy (i) so that $f-g$ is convex but $f$ and $g$ are individually concave.
When the domain is $\mathbb{R}$ and $f$ and $g$ are both twice differentiable, it is also sufficient to have
$$
f''\geq g''\tag{ii}.
$$
This is false as stated. Quasiconvexity definition says that the lower level sets $\{f\le \lambda\}$ are convex. Unimodularity says that $f$ increases along the rays emanating from its global minimum. The first certainly implies the second. But the converse is false.
Example: $f(x_1,x_2) = \sqrt{|x_1|}+\sqrt{|x_2|}$. Clearly, $-f$ is unimodal with global maximum at $(0,0)$. But the lower level sets are not convex: they are bounded by astroids.
Concretely, $f(1/2,1/2) = \sqrt{2}$ is greater than $f(1,0)=f(0,1)=1$.
Best Answer
Yes a generalization is possible. Here is an elementary approach to the convexity of the product of two nonnegative convex functions defined on a convex domain of $\mathbb{R}^n$.
Choose $x$ and $y$ in the domain and $t$ in $[0,1]$. Your aim is to prove that $\Delta\ge0$ with $$\Delta=t(fg)(x)+(1-t)(fg)(y)-(fg)(tx+(1-t)y). $$ But $f$ and $g$ are nonnegative and convex, hence $$ (fg)(tx+(1-t)y)\le(tf(x)+(1-t)f(y))(tg(x)+(1-t)g(y)). $$ Using this and some easy algebraic manipulations, one sees that $\Delta\ge t(1-t)D(x,y)$ with $$ D(x,y)=f(x)g(x)+f(y)g(y)-f(x)g(y)-f(y)g(x), $$ that is, $$ D(x,y)=(f(x)-f(y))(g(x)-g(y)). $$ This proves a generalization of the result you quoted to any product of convex nonnegative functions $f$ and $g$ such that $D(x,y)\ge0$ for every $x$ and $y$ in the domain of $f$ and $g$.
In particular, if $f$ and $g$ are differentiable at a point $x$ in the domain, one asks that their gradients $\nabla f(x)$ and $\nabla g(x)$ are such that $z^*M(x)z\ge0$ for every $n\times 1$ vector $z$, where $M(x)$ is the $n\times n$ matrix $$ M(x)=\nabla f(x)\cdot(\nabla g(x))^*. $$