While the question has been answered, it is interesting to check what happens if we slightly change the condition, from $g(x+s)-g(x)<t$ to $g(x+s)-g(x)\leq t$. [The above proof][1] does not work. 
However, I still think that it is impossible to decide if such $x$ exists with finitely-many queries. Fixing $s$ and $t$, for every $z\in[0,1-s]$, let $G_z$ be the set of continuous functions $g_z$ for which:

$$
g_z(x+s) - g_z(x) > t ~~~~ x\neq z
\\
g_z(x+s) - g_z(x) = t ~~~~ x = z
$$

(it should be possible to construct such continuous functions; I do not have the exact construction now).

To prove impossibility, we can use an adversary argument: we show that, for any algorithm for asking adaptive queries, an adversary can answer the queries in such a way that the algorithm will never know if a solution exists or not.

The adversary works as follows: he picks an arbitrary $z\in[0,1-s]$, and an arbitrary $g_z\in G_z$, and answers all queries as if $g \equiv g_z$, as long as the queries do not involve the point $z$ itself. In case a query does involve the point $z$, the adversary picks a nearby point $z'$, that is not equal to any recorded point (any point that appeared in a previous query). He constructs a new function $g_{z'}\in G_{z'}$, that coincides with $g_z$ in all recorded points (there are finitely many such points, so it should be possible to construct such a continuous function). The adversary can keep switching functions forever, and the algorithm will never know the actual $z$, and thus will never know if a solution exists.

  [1]: https://mathoverflow.net/a/371325/34461