Ok, I think the answer is no. There does not exist a utility function satisfying DARA such that $A(x)>\frac{1}{x}$ for all $x\in R_+$.
To prove that, let's assume such a function exists. Then, there must exists $a\in R_+$ such that:
$$\frac{-f''}{f'}>\frac{1}{x}\Leftrightarrow\frac{-f''}{f'}-\frac{1}{x}-a=0$$
If we rearrange this function a bit by taking $y=x+w$, we have:
$$-\frac{f''(y)}{f'(y)}=\frac{1+a(y-w)}{y-w}\qquad\text{(1)}$$
Expression (1) is a second-order nonlinear ordinary differential equation. Solving it (here pardon my laziness, I used Wolfram) gives:
$$f(y)=\int_1^{y}exp\bigg(\int_{1}^{\phi}\frac{a(\rho-w)+1}{w-\rho}d\rho\bigg)c_1d\phi+c_2$$
Taking the derivative of $f$ we get:
$$f'(y)=exp\bigg(\int_1^{y}\frac{a(\rho-w)+1}{w-\rho}d\rho\bigg)c_1$$
So we do have the condition $f'>0$ fulfilled as long as $c_1>0$. Taking the second-order derivative of $f$, we get:
$$f''(y)=\frac{a(y-w)+1}{w-y}exp\bigg(\int_1^{y}\frac{a(\rho-w)+1}{w-\rho}d\rho\bigg)c_1$$
This expression is negative if and only if:
$$\frac{a(y-w)+1}{w-y}>0$$
Using the fact that $x=y-w$:
$$-\frac{1+ax}{x}>0\Leftrightarrow x<-\frac{1}{a}$$
We have a contradiction -- if I did not make any mistake in the computations -- there does not exist a DARA utility function where $A(x)>\frac{1}{x}$, for all $x\in R_+$.
