If you are allowed to use one extra variable, then a queue can simulate a stack, so in this model a queue plus a single extra storage location will suffice.
The idea is to represent a stack, say $[\ a\ b\ c\ $(top), as a queue, (tail) $\langle c\ b\ a\rangle$ (head). Then the operation $\mathtt{push(x)}$ would correspond to $\mathtt{enque(x)}$, so
$$
[\ a\ b\ c\quad \stackrel{\mathtt{push(x)}}{\longrightarrow}\quad[\ a\ b\ c\ x
$$
would be simulated as
$$
\langle\ c\ b\ a \rangle\quad \stackrel{\mathtt{enque(x)}}{\longrightarrow}\quad\langle\ x\ c\ b\ a\ \rangle
$$
The $\mathtt{pop()}$ operation is a bit more complicated. We first enqueue a marker, #, and then keep pulling elements off the head of the queue and placing them back on the tail until we come to the marker, indicating that the last element we pulled off corresponds to the top of the stack, so we don’t replace that element on the queue and we finish by removing the marker. In pseudocode, what we do is:
enque(#)
v = head() ; save the head element
deque() ; remove it
while (head() != #)
enque(v) ; put the element back on the tail
v = head() ; save the next element
deque() ; remove it
deque() ; finally, remove the marker
For example, starting with $\langle c\ b\ a\rangle$, we'd have
$$\begin{align}
\langle\ c\ b\ a\ \rangle &\longrightarrow \langle\ \mathtt{\#}\ c\ b\ a\ \rangle \\
&\longrightarrow \langle\ a\ \mathtt{\#}\ c\ b\ \rangle \\
&\longrightarrow \langle\ b\ a\ \mathtt{\#}\ c\ \rangle \\
&\longrightarrow \langle\ b\ a\ \mathtt{\#}\ \rangle \\
&\longrightarrow \langle\ b\ a\ \rangle
\end{align}$$