To find the behavior when $c \to 0$ we'll split the integral into the two pieces
$$
\int_{0}^{\infty} dk\, e^{-ak^2} J_0(bk) \frac{k^3}{c^2+k^4} = \left( \int_{0}^{\sqrt{c}} + \int_{\sqrt{c}}^{\infty} \right) dk\, e^{-ak^2} J_0(bk) \frac{k^3}{c^2+k^4}. \tag{1}
$$
The first piece can be expanded in powers of $c$ by writing
$$
e^{-ak^2} J_0(bk) = \sum_{j=0}^{\infty} \alpha_j k^{2j},
$$
then substituting this in to find that
$$
\begin{align}
&\int_{0}^{\sqrt{c}} dk\, e^{-ak^2} J_0(bk) \frac{k^3}{c^2+k^4} \\
&\qquad = \sum_{j=0}^{\infty} \alpha_j \int_0^\sqrt{c} \frac{k^{3+2j}}{c^2+k^4}\,dk \\
&\qquad = \sum_{j=0}^{\infty} \left(\alpha_j \int_0^1 dx\, \frac{x^{3+2j}}{1+x^4} \right) c^j \\
&\qquad = \frac{1}{4}\log 2 - \frac{(4-\pi)(4a-b^2)}{32} c + \frac{(1-\log 2)(32 a^2+16 a b^2+b^4)}{256} c^2 + \cdots,
\end{align}
$$
$$
\tag{2}
$$
so it stands to reason that the leading-order asymptotic behavior of the integral comes from the second piece. Let's use the fact that
$$
\frac{k^3}{c^2+k^4} = \frac{1}{k} - \frac{c^2}{k(c^2+k^4)}
$$
to write it as
$$
\int_{\sqrt{c}}^{\infty} dk\, e^{-ak^2} J_0(bk) \frac{k^3}{c^2+k^4} = \int_{\sqrt{c}}^{\infty} dk\, e^{-ak^2} J_0(bk) k^{-1} - c^2 \int_{\sqrt{c}}^{\infty} dk\, e^{-ak^2} J_0(bk) \frac{1}{k(c^2+k^4)}
$$
$$
\tag{3}
$$
Fix $0 < \epsilon < 1/2$ and split the second integral like
$$
c^2 \int_{\sqrt{c}}^{\infty} dk\, e^{-ak^2} J_0(bk) \frac{1}{k(c^2+k^4)} = c^2 \left(\int_{\sqrt{c}}^{\Large c^\epsilon} + \int_{\Large c^\epsilon}^{\infty} \right) dk\, e^{-ak^2} J_0(bk) \frac{1}{k(c^2+k^4)}.
$$
$$
\tag{4}
$$
The tail term is bounded by
$$
\left| c^2 \int_{\Large c^\epsilon}^{\infty} dk\, e^{-ak^2} J_0(bk) \frac{1}{k(c^2+k^4)} \right| < c^2 \int_{\Large c^\epsilon}^\infty \frac{dk}{k (c^2+k^4)} = \frac{1}{4} \log(1 + c^{2-4\epsilon}) \tag{5}
$$
and the first can be expanded as before;
$$
c^2 \int_{\sqrt{c}}^{\Large c^\epsilon} dk\, e^{-ak^2} J_0(bk) \frac{1}{k(c^2+k^4)} = c^2 \sum_{j=0}^{\infty} \alpha_j \int_{\sqrt{c}}^{\Large c^\epsilon} dk\, \frac{k^{2j}}{k(c^2+k^4)}.
$$
In fact we'll show that all we can use are the first two terms of this expansion, so for now we'll just write
$$
\begin{align}
&c^2 \int_{\sqrt{c}}^{\Large c^\epsilon} dk\, e^{-ak^2} J_0(bk) \frac{1}{k(c^2+k^4)} \\
&\qquad = c^2 \int_\sqrt{c}^{\Large c^\epsilon} \frac{dk}{k (c^2+k^4)} - \left(a+\frac{b^2}{4}\right) c^2 \int_\sqrt{c}^{\Large c^\epsilon} dk\, \frac{k}{c^2+k^4} + O\left(c^2 \int_\sqrt{c}^{\Large c^\epsilon} dk\, \frac{k^3}{c^2+k^4}\right).
\end{align}
$$
$$
\tag{6}
$$
We must now estimate these new integrals. For the first we make the change of variables $k = \sqrt{c} x$ to get
$$
c^2 \int_\sqrt{c}^{\Large c^\epsilon} \frac{dk}{k (c^2+k^4)} = \int_1^{\Large c^{\epsilon - 1/2}} \frac{dx}{x (1+x^4)} = \int_1^\infty \frac{dx}{x (1+x^4)} - \int_{\Large c^{\epsilon-1/2}}^\infty \frac{dx}{x (1+x^4)}.
$$
Of course
$$
\int_1^\infty \frac{dx}{x (1+x^4)} = \frac{1}{4}\log 2
$$
and
$$
0 < \int_{\Large c^{\epsilon-1/2}}^\infty \frac{dx}{x (1+x^4)} < \int_{\Large c^{\epsilon-1/2}}^\infty \frac{dx}{x^5} = \frac{1}{4} c^{2-4\epsilon},
$$
so we just end up with
$$
c^2 \int_\sqrt{c}^{\Large c^\epsilon} \frac{dk}{k (c^2+k^4)} = \frac{1}{4}\log 2 + O(c^{2-4\epsilon}). \tag{7}
$$
An identical argument applied to the second integral yields
$$
c^2 \int_\sqrt{c}^{\Large c^\epsilon} dk\, \frac{k}{c^2+k^4} = \frac{\pi}{8} c + O(c^{2-4\epsilon}). \tag{8}
$$
For the last integral we only need the blunt estimate
$$
0 < c^2 \int_\sqrt{c}^{\Large c^\epsilon} dk\, \frac{k^3}{c^2+k^4} = c^2 \int_1^{\Large c^{\epsilon-1/2}} \frac{x^3}{1+x^4} < c^2 \log c^{\epsilon-1/2} < c^{2-4\epsilon}
$$
for $c$ small enough. By combining this with $(7)$ and $(6)$ in $(5)$ we get
$$
c^2 \int_{\sqrt{c}}^{\Large c^\epsilon} dk\, e^{-ak^2} J_0(bk) \frac{1}{k(c^2+k^4)} = \frac{1}{4}\log 2 - \frac{\pi}{8} \left(a+\frac{b^2}{4}\right) c + O(c^{2-4\epsilon}), \tag{9}
$$
and this, combined with $(5)$ in $(4)$, yields
$$
c^2 \int_{\sqrt{c}}^{\infty} dk\, e^{-ak^2} J_0(bk) \frac{1}{k(c^2+k^4)} = \frac{1}{4}\log 2 - \frac{\pi}{8} \left(a+\frac{b^2}{4}\right) c + O(c^{2-4\epsilon}). \tag{10}
$$
Thus $(3)$ becomes
$$
\begin{align}
&\int_{\sqrt{c}}^{\infty} dk\, e^{-ak^2} J_0(bk) \frac{k^3}{c^2+k^4} \\
&\qquad = \int_{\sqrt{c}}^{\infty} dk\, e^{-ak^2} J_0(bk) k^{-1} - \frac{1}{4}\log 2 + \frac{\pi}{8} \left(a+\frac{b^2}{4}\right) c + O(c^{2-\epsilon}) \tag{11}
\end{align}
$$
as $c \to 0^+$ for any fixed $\epsilon > 0$. Finally, we can write the integral here as
$$
\begin{align}
&\int_{\sqrt{c}}^{\infty} dk\, e^{-ak^2} J_0(bk) k^{-1} \\
&\qquad = \int_{\sqrt{c}}^{\infty} dk\, e^{-k} k^{-1} + \int_{\sqrt{c}}^{\infty} dk\, \Bigl( e^{-ak^2} J_0(bk) - e^{-k} \Bigr) k^{-1} \\
&\qquad = -\operatorname{Ei}\left(-\sqrt{c}\right) + \int_{\sqrt{c}}^{\infty} dk\, \Bigl( e^{-ak^2} J_0(bk) - e^{-k} \Bigr) k^{-1},
\end{align}
$$
where $\operatorname{Ei}$ is the exponential integral. Now
$$
\begin{align}
&\int_{\sqrt{c}}^{\infty} dk\, \Bigl( e^{-ak^2} J_0(bk) - e^{-k} \Bigr) k^{-1} \\
&\qquad = \int_0^\infty dk\, \Bigl( e^{-ak^2} J_0(bk) - e^{-k} \Bigr) k^{-1} - \int_0^\sqrt{c} dk\, \Bigl( e^{-ak^2} J_0(bk) - e^{-k} \Bigr) k^{-1} \\
&\qquad = f(a,b) - \sum_{j=0}^{2} \beta_j \int_0^\sqrt{c} dk\, k^j + O(c^2) \\
&\qquad = f(a,b) - \sqrt{c} + \frac{1}{2}\left(a + \frac{b^2}{4} + \frac{1}{2}\right) c - \frac{1}{18} c^{3/2} + O(c^2),
\end{align}
$$
where
$$
f(a,b) := \int_0^\infty dk\, \Bigl( e^{-ak^2} J_0(bk) - e^{-k} \Bigr) k^{-1}
$$
and the coefficients $\beta_j$ are defined by
$$
\begin{align}
\Bigl( e^{-ak^2} J_0(bk) - e^{-k} \Bigr) k^{-1} &= \sum_{j=0}^{\infty} \beta_j k^j \\
&= 1 - \left(a + \frac{b^2}{4} + \frac{1}{2}\right) k + \frac{1}{6} k^2 + \cdots,
\end{align}
$$
so that
$$
\begin{align}
&\int_{\sqrt{c}}^{\infty} dk\, e^{-ak^2} J_0(bk) k^{-1} \\
&\qquad = -\operatorname{Ei}\left(-\sqrt{c}\right) + f(a,b) - \sqrt{c} + \frac{1}{2}\left(a + \frac{b^2}{4} + \frac{1}{2}\right) c - \frac{1}{18} c^{3/2} + O(c^2).
\end{align}
$$
Substituting this into $(11)$ thus yields
$$
\begin{align}
&\int_{\sqrt{c}}^{\infty} dk\, e^{-ak^2} J_0(bk) \frac{k^3}{c^2+k^4} \\
&\qquad = -\operatorname{Ei}\left(-\sqrt{c}\right) + f(a,b) - \frac{1}{4}\log 2 - \sqrt{c} + \left( \frac{1}{4} + \frac{\pi+4}{8}a + \frac{\pi+4}{32} b^2\right) c \\
&\qquad \qquad - \frac{1}{18}c^{3/2} + O(c^{2-\epsilon}),
\end{align}
$$
$$
\tag{12}
$$
and, at last, combining this with $(2)$ in $(1)$ grants us
$$
\begin{align}
&\int_{0}^{\infty} dk\, e^{-ak^2} J_0(bk) \frac{k^3}{c^2+k^4} \\
&\qquad = -\operatorname{Ei}\left(-\sqrt{c}\right) + f(a,b) - \sqrt{c} + \frac{1 + \pi a + b^2}{4} c - \frac{1}{18} c^{3/2} + O(c^{2-\epsilon}). \tag{13}
\end{align}
$$
It is known (see wikipedia) that
$$
\operatorname{Ei}(z) = \log|z| + \gamma + x + \frac{1}{4}x^2 + \frac{1}{18}x^3 + O(x^4)
$$
as $x \to 0$, so in our case we have
$$
-\operatorname{Ei}\left(-\sqrt{c}\right) = \frac{1}{2} \log \frac{1}{c} - \gamma + \sqrt{c} - \frac{1}{4} c + \frac{1}{18} c^{3/2} + O(c^2),
$$
and so we arrive at the asymptotic
$$
\begin{align}
&\int_{0}^{\infty} dk\, e^{-ak^2} J_0(bk) \frac{k^3}{c^2+k^4} \\
&\qquad = \frac{1}{2} \log \frac{1}{|c|} + f(a,b) - \gamma + \frac{\pi a+b^2}{4} |c| + O(|c|^{2-\epsilon}) \tag{14}
\end{align}
$$
as $c \to 0$ for any fixed $\epsilon > 0$, where
$$
f(a,b) = \int_0^\infty dk\, \Bigl( e^{-ak^2} J_0(bk) - e^{-k} \Bigr) k^{-1}.
$$
Surely the $O(|c|^{2-\epsilon})$ term may be replaced with $\Theta(c^2 \log |c|)$ with a little more work.
Here's a log-log plot to illustrate the asymptotic with $a=b=1$ over the range $c \in (2\cdot 10^{-4},10^{-2})$. The black points are numerical evaluations of the given integral and the blue curve is
$$
\frac{1}{2} \log \frac{1}{c} + f(1,1) - \gamma.
$$
