Here’s another perspective, by far not as elegant as the one in lulu’s answer, but perhaps it throws some more light on what misleads our intuition on this nice problem.
A quantitative form of the argument for the wrong intuition that the jack should be more likely would be: The probability that $Q\heartsuit$ has already been drawn after the first queen is drawn is $\frac14$, whereas the probability that $J\spadesuit$ has already been drawn is $\frac15$ (since that card and the four queens are all equally likely to be drawn first). Thus it’s more likely that $J\spadesuit$ is still in the deck than that $Q\heartsuit$ is still in the deck.
The mistake here is that these are just the marginal probabilities, but these events correlate with how many cards are still in the deck, and while $J\spadesuit$ is more likely to still be in the deck, in that case the deck is more likely to still contain more cards, which makes it less likely that $J\spadesuit$ will be drawn next.
The probability that, after the first queen is drawn, $Q\heartsuit$ has been drawn and there are $k$ cards left is
$$
\mathsf{Pr}(Q\heartsuit\text{ drawn}\land K=k)=\frac14\frac{\binom k3}{\binom{52}4}\;,
$$
since there are $\binom{52}4$ ways to choose where the $4$ queens are, and for the event to occur $Q\heartsuit$ has to be in a particular place and the remaining $3$ queens must be among the remaining $k$ cards.
The probability that, after the first queen is drawn, $J\spadesuit$ has been drawn and there are $k$ cards left is
$$
\mathsf{Pr}(J\spadesuit\text{ drawn}\land K=k)=\frac{51-k}{48}\frac{\binom k3}{\binom{52}4}\;,
$$
since one queen must be in a particular place, the other three must be among the remaining $k$ cards and $J\spadesuit$ must be among the $51-k$ cards drawn before the first queen, and there are $48$ places left where it could be.
Summing these over $k$ yields $\frac14$ and $\frac15$, respectively, as it must. But to get the probability that $Q\heartsuit$ or $J\spadesuit$ is drawn next, we need to sum the probability that that card hasn’t been drawn times $\frac1k$:
\begin{eqnarray}
\mathsf{Pr}(Q\heartsuit\text{ is next})
&=&
\sum_{k=3}^{51}\mathsf{Pr}(Q\heartsuit\text{ not drawn}\land K=k)\cdot\frac1k
\\
&=&
\sum_{k=3}^{51}\left(1-\frac14\right)\frac{\binom k3}{\binom{52}4}\cdot\frac1k
\\
&=&
\frac1{52}
\end{eqnarray}
and
\begin{eqnarray}
\mathsf{Pr}(J\spadesuit\text{ is next})
&=&
\sum_{k=3}^{51}\mathsf{Pr}(J\spadesuit\text{ not drawn}\land K=k)\cdot\frac1k
\\
&=&
\sum_{k=3}^{51}\left(1-\frac{51-k}{48}\right)\frac{\binom k3}{\binom{52}4}\cdot\frac1k
\\
&=&
\frac1{52}\;.
\end{eqnarray}
Here’s a Wolfram|Alpha plot of $\mathsf{Pr}(Q\heartsuit\text{ not drawn}\land K=k)$ and $\mathsf{Pr}(J\spadesuit\text{ not drawn}\land K=k)$ over $k$. You can see that the area under the curve for $J\spadesuit$ is a bit larger, but more of that area is at high $k$, whereas the probability for $Q\heartsuit$ is higher than the one for $J\spadesuit$ for $k\lt39$. The expected value of $k$ given that $Q\heartsuit$ or $J\spadesuit$ is still in the deck is
\begin{eqnarray}
\mathsf E[K\mid Q\heartsuit\text{ not drawn}]
&=&
\frac43\sum_{k=3}^{51}\mathsf{Pr}(Q\heartsuit\text{ not drawn}\land K=k)\cdot k
\\
&=&
\frac43\sum_{k=3}^{51}\left(1-\frac14\right)\frac{\binom k3}{\binom{52}4}\cdot k
\\
&=&
\frac{207}5
\\
&=&
41.4
\end{eqnarray}
and
\begin{eqnarray}
\mathsf E[K\mid J\spadesuit\text{ not drawn}]
&=&
\frac54\sum_{k=3}^{51}\mathsf{Pr}(J\spadesuit\text{ not drawn}\land K=k)\cdot k
\\
&=&
\frac54\sum_{k=3}^{51}\left(1-\frac{51-k}{48}\right)\frac{\binom k3}{\binom{52}4}\cdot k
\\
&=&\frac{259}6
\\
&=&43.1\overline6\;,
\end{eqnarray}
respectively, not that much of a difference, but what matters is the probabilities for small values of $k$, where it’s most likely to actually draw the card next if it’s still there, and since the ratio of the two probabilities is $\frac{k-3}{36}$, the queen has a considerable advantage at small $k$. This, of course, goes back to what’s already been discussed in comments, that $J\spadesuit$ can be gone due to having been drawn as any of the cards before the first queen, whereas $Q\heartsuit$ can only be gone due to being the first queen, which always has the same probability of $\frac14$ no matter how many cards have already been drawn.
