It is known (see Titchmarsh Chapter 12) that if you define $\gamma_k$ the lower bound of $\sigma > 0, \int_{-\infty}^{\infty}\frac{|\zeta(\sigma+it)|^{2k}}{|\sigma+it|^2}dt < \infty$, then $\frac{k-1}{2k}\le \gamma_k=\beta_k \le \alpha_k$, where $\beta_k$ is the usual mean bound of $\Delta_k$ ( lowest bound of orders for which $\frac{1}{x}\int_{0}^{x}{\Delta_k(y)^2}dy=O_{\epsilon}(x^{2\beta_k+\epsilon})$) and the integral above in the question converges to $\Delta_k(x)$ for $\gamma_k< \gamma <1$.
Since it is known that for example $\gamma_2=\frac{1}{4}, \gamma_3=\frac{1}{3}, \gamma_4=\frac{3}{8}$ (best possible, first two results appearing in Titchmarsh, last in Ivic), while the known respective values for $\alpha$ are still far away from those, even in the simplest cases $k=2,3,4$ the equivalence required is far from being proven and it is equivalent to fully solving the Dirichlet Divisor problem for $k=2,3,4$.
In general, obviously, even less is known about $\gamma_k, \alpha_k$ so the question is on par with Lindelof in many ways (as Lindelof is equivalent to any and all of $\alpha_k \le \frac{1}{2}, \gamma_k \le \frac{1}{2}, \gamma_k = \frac{k-1}{2k}$)