Yes it does, the Generalized Beta prime distribution with shape parameter equal to 1.
We can get there fairly easily by integrating $\beta$ out of the joint distribution of $x$ and $\beta$:
$$f(x,\beta|\alpha, \alpha_0, \beta_0) = {\beta^{\alpha}x^{\alpha-1} \over \Gamma(\alpha)}e^{-\beta x}{\beta_0^{\alpha_0}\beta^{\alpha_0-1} \over \Gamma{\alpha_0}}e^{-\beta_0\beta}$$
Rearranging terms and ignoring everything that isn't related to either $\beta$ or $x$ (as they will all be handled by the constant of integration at the end) results in a great deal of simplification:
$$f(x,\beta|\cdot) \propto x^{\alpha-1}\beta^{\alpha+\alpha_0-1}e^{-(\beta_0+x)\beta}$$
We can integrate out $\beta$ easily enough by noting that the two terms involving $\beta$ are those of a Gamma-distributed variate with shape parameter $\alpha + \alpha_0$ and rate parameter $\beta_0 + x$, so the integral must equal the inverse of the constant of integration of such a distribution:
$$x^{\alpha-1}\int \beta^{\alpha+\alpha_0-1}e^{-(\beta_0+x)\beta}\text{d}\beta = {x^{\alpha-1} \Gamma(\alpha+\alpha_0) \over(\beta_0 + x)^{\alpha + \alpha_0}} \propto x^{\alpha-1} (\beta_0 + x)^{-\alpha - \alpha_0}$$
A slight rearrangement of terms and some minor algebra gets us to:
$$f(x|\cdot) \propto \left({x \over \beta_0}\right)^{\alpha-1} \left(1 + {x\over \beta_0}\right)^{-\alpha - \alpha_0}$$
which clearly matches the formula for the GBPD
(with shape parameter $p=1$) as given by Wikipedia and reproduced here:
$$f(x;\alpha,\beta,p,q) = \frac{p \left(\frac x q \right)^{\alpha p-1} \left(1+ \left(\frac x q \right)^p\right)^{-\alpha -\beta}}{qB(\alpha,\beta)}$$