I have arrived at an elementary-looking "result" via a sketchy argument. Having unsuccessfully searched for the statement and its "proof" in the literature, I would like to hear if anyone knows whether or not it is true, and if they might provide me with a reference? The statement is as follows: let $f:\mathbb{N}\rightarrow \mathbb{C}$ and
$$F(s)=\sum_{1}^{\infty}\frac{f(n)}{n^s}$$
converge for $\sigma>\sigma_c>0$, then
$$\int_{\sigma-i\infty}^{\sigma+i\infty}|F(s)|^2\frac{ds}{s}$$
converges for $\sigma>\sigma_c>0$, and vice-versa.
EDIT 1: "Vice-versa" meaning that the converse implication also holds.
EDIT 2: The integrand is to be taken as an analytic continuation of the series definition of $F$, assuming a-priori that one exists there.
Sketch of proof: I begin by constructing an auxillary Dirichlet series:
$$D(s)=\sum_{1}^{\infty}\frac{d(n)}{n^s}$$
where
$$d(n)=2\left(f(1)+f(2)+\cdots+\frac{f(n)}{2}\right)\bar{f}(n).$$
Thus
$$\sum_{1}^{n}d(m)=\left|\sum_{1}^{n}f(m)\right|^2$$
is positive. The abscissa of convergence of $D(s)$ is
$$\sigma_d=\limsup_{n\rightarrow\infty}\frac{\log\sum_{1}^{n}d(m)}{\log n}=2\limsup_{n\rightarrow\infty}\frac{\log\left|\sum_{1}^{n}f(m)\right|}{\log n}=2\sigma_c,$$
provided $\sigma_c>0$. I argue that the convergence of the Dirichlet series $D(2s)$ is equivalent to the convergence of the integral above as follows: If $\sigma>\sigma_c$, then
$$D(2\sigma)=\sum_{1}^{\infty}\frac{d(n)}{n^{2\sigma}}=\sum_{1}^{\infty}\left(\frac{1}{\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}\sum_{1}^{\infty}\frac{f(m)}{m^{s}}\frac{n^{s}ds}{s}\right)\frac{\bar{f}(n)}{n^{2\sigma}},$$
where both $D(2s)$ and $F(s)$ converge absolutely if $\sigma>\sigma_c+1$. Since $F(s)$ has no poles for $\sigma>\sigma_c>0$, so for fixed $n$ the (conditionally convergent) integral is unchanged for $\sigma>\sigma_c$. If one can justify interchanging the order of summation and integration, we get
$$D(2\sigma)=\frac{1}{\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}|F(s)|^2\frac{ds}{s}.$$
Best Answer
Something similar is true. Let $$ A(x):=\sum_{n\leq x}f(n),\qquad x>0,$$ then the first condition is equivalent to $$ \forall\sigma>\sigma_c:\ A(x)\ll_\sigma x^\sigma. \tag{1}$$ This clearly implies $$ \forall\sigma>\sigma_c:\ \int_0^\infty |A(x)|^2 x^{-2\sigma}\frac{dx}{x}<\infty, \tag{2}$$ which by $$ F(s)=s\int_1^\infty A(x)x^{-s}\frac{dx}{x},\qquad\Re s>\sigma_c, $$ is equivalent to $$ \forall\sigma>\sigma_c:\ \int_{\sigma-i\infty}^{\sigma+i\infty}|F(s)|^2\frac{ds}{|s|^2}<\infty. \tag{3}$$ This suggests that your second condition should be changed to (3).
Still, the converse (3)=>(1) would be false, for it might well happen for a step function $A(x)$ that (2) holds but (1) fails. On the other hand, (2) and hence also (3) express the fact that (1) holds for most $x$'s.