This is not a complete answer
The argument below doesn't give boundedness.
However, it gives that under the assumption $\int_1^\infty x(s) k(s) s^{-2} ds$ converges, $x(t) = o(t^\lambda)$ for any $\lambda > 0$.
With stronger assumptions on $k$ the argument can also imply boundedness.
Sublinear growth
First one observes that if the integral is finite, the growth rate is sublinear. This follows from the fact that if the integral $\int_1^\infty x(s) k(s) s^{-2} ds$ converges, then $x'(t) \to 0$ as $t\to\infty$, and hence $x(t) = o(t)$ by, e.g., L'Hopital.
Upgrade the growth control
Integrating the bound for $x'(t)$ gives, for arbitrary $1 \leq t_1 < t_2$
$$ x(t_2) - x(t_1) \leq \int_{t_1}^{t_2} \int_t^\infty x(s) k(s) s^{-2} ds~dt $$
Integrating in parts in $t$, using that $dt = d(t - t_1)$
$$ x(t_2) - x(t_1) \leq (t_2 - t_1) \int_{t_2}^\infty x(s) k(s) s^{-2} ds + \int_{t_1}^{t_2} (t - t_1) x(t) k(t) t^{-2} dt $$
Now, for any $\delta, \beta> 0$, there exists $\tau_{\delta,\beta}$ such that for every $s \geq \tau_{\delta,\beta}$, it holds that
$$ x(s) \leq \delta s, \qquad \text{and} \qquad \int_{s}^\infty k(s)s^{-1} ds < \beta $$
since we know that $x$ grows sublinearly and the relevant integral converges. This implies that, for $t_2 > \tau_{\delta,\beta}$,
$$ x(t_2) \leq x(\tau_{\delta,\beta}) + (t_2 - \tau_{\delta,\beta}) \delta \beta + \int_{\tau_{\delta,\beta}}^{t_2} x(s) \cdot k(s) s^{-1} ds $$
By Gronwall's inequality this means that
$$ x(t_2) \leq [ \delta \tau_{\delta,\beta} + (t_2 - \tau_{\delta}) \delta \beta ] e^{\beta} $$
This final inequality can be used to estimate $\tau_{\delta/2,\beta}$. Consider the inequality
$$ [\delta \tau_{\delta,\beta} + (s - \tau_{\delta,\beta}) \delta\beta] e^{\beta} \leq \frac{\delta}{2} s \tag{A}$$
Solving this inequality we see that this is satisfied whenever
$$ \frac{e^{\beta} \tau_{\delta,\beta}}{\frac12 - \beta e^{\beta}} \leq s $$
For convenience write
$$ \frac{e^\beta}{\frac12 - \beta e^\beta} = 2^{1+\sigma(\beta)} $$
and note that $\sigma$ is continuous and $\sigma(0) = 0$.
This implies that $\tau_{\delta/2,\beta} \leq 2^{1 + \sigma(\beta)} \tau_{\delta}$.
Now, iterating this argument, starting from some sufficiently small $\delta$ that we fix, we see can construct an increasing sequence of times $\tau_{2^{-k} \delta,\beta}$ such that for $T \geq \tau_{2^{-k} \delta,\beta}$ it holds
$$ x(T) \leq 2^{-k} \delta T $$
We also have by iteration the bounds
$$ \tau_{2^{-k}\delta,\beta} \leq 2^{k + k\sigma(\beta)} \tau_{\delta,\beta} $$
So we have
$$ x(T) \lesssim_{\delta,\beta} T^{\sigma(\beta)} $$
Boundedness
One can upgrade the estimate to get boundedness if one knows the decay rate of $\int_t^\infty k(s) s^{-1} ds$.
For example, if one knows that this integral is bounded by $C/\ln(t)^2$, then Gronwall will imply an estimate along the lines of
$$ (\tau_{\delta} + (s - \tau_\delta) \frac{C}{(\ln \tau_{\delta})^2}) e^{C/(\ln \tau_\delta)^2} < \frac12 s $$
replacing (A).
Now, by our estimates in the previous section we would've found that
$$ \tau_{2^{-k\delta}} \leq 2^{k(1 + \sigma)} \tau_{\delta} $$
for some $\sigma$. Bootstrapping from this we would get that
$$ \tau_{2^{-k} \delta} \leq 2^{1 + O(k^{-2})} \tau_{2^{-(k-1)} \delta} $$
The summability of the series $(k^{-2})$ then gives boundedness of $x(t)$.
[This argument can carry through as long as we know, for example, that $k(s) \lesssim (\ln (1+s))^{-2-\gamma}$ for some $\gamma > 0$; but seemingly fails for $k(s) = (\ln(1+s))^{-2}$.]