Both $6_1$ and $3_1 \# m(3_1)$ are smoothly slice (as is the unknot), and I claim that all other knots of at most seven crossings are not integrally slice. This will follow from two claims: first, if $K$ is integrally slice then $\frac{\Delta_K''(1)}{2}$ is even; and second, if $K$ is alternating and rationally slice then its signature is zero. (Note that $8_3$ satisfies both of these but is not smoothly slice; I don't know whether it is integrally slice.)
The key observation is that if $K$ is integrally slice, then $S^3_1(K)$ must be integrally homology cobordant to $S^3$. To see this, we take a slice disk bounded by $K$ in a homology ball, and we remove a ball about some point on the disk to get a concordance from $U$ to $K$ inside a homology cobordism from $S^3$ to itself. Performing a 1-surgery along this cylinder gives us the desired homology cobordism.
From here, filling in the $S^3$ end with a ball gives us a smooth homology ball bounded by $S^3_1(K)$. Thus $S^3_1(K)$ has vanishing Rohklin invariant, or equivalently its Casson invariant is even, and the surgery formula for the latter says that $\frac{\Delta_K''(1)}{2}$ must be even.
For the second claim, $S^3_1(K)$ is rationally homology cobordant to $S^3$, so its Heegaard Floer d-invariant must be zero. When $K$ is alternating, this was computed to be $2\min(0,-\lceil -\sigma(K)/4\rceil)$ for alternating $K$ by Ozsváth and Szabó (arXiv:0209149, corollary 1.5), so we must have $\sigma(K) \geq 0$. But the same argument applies to the mirror $m(K)$, with signature $-\sigma(K)$, so in fact $K$ must have signature zero.
