Are there languages in $DTIME(2^{t(n)})$ that are not in $NTIME(t(n))$?
If $\mathrm{NTIME}(n) \supseteq \mathrm{DTIME}(2^{n/2000})$ (where NTIME and DTIME uses constant factor in the big-oh) then for all time-constructible $t\geq n$, we have that $\mathrm{NTIME}(t) = \mathrm{DTIME}(2^{\mathrm{O}(t)})\supsetneq \mathrm{DTIME}(2^t)$ to the contrary to your quest. So, in that case, requiring the NTM to read a tape of nondeterministic guesses are essential to separate these classes.

And it is still open that $\mathrm{NTIME}(n) = \mathrm{E}$.



Given a language $\mathrm{L}\in \mathrm{DTIME}(2^{\mathrm{O}(t)})$ decided by a TM running in time $2^{ct}$ where $c$ is some constant, construct the language $\mathrm{L}_\mathrm{pad} = \{x10^{2000*ct-1-\vert x\vert}\vert x\in \mathrm{L}\}$. Clearly, $\mathrm{L}_\mathrm{pad}$ can be decided by a TM running in time $\mathrm{DTIME}(2^{n/2000})$, i.e. $\mathrm{L}_\mathrm{pad}\in \mathrm{DTIME}(2^{n/2000})$. By our assumption, we have that $\mathrm{L}_\mathrm{pad}\in \mathrm{NTIME}(n)$, i.e. there exists an NTM deciding $\mathrm{L}_\mathrm{pad}$ in non-deterministic time $dn$ for some constant $d$. Now, we describe the NTM for $\mathrm{L}$. When given an input dtring $x$, our NTM for $\mathrm{L}$ first pads it up to length $2000*ct$ and then run the $\mathrm{NTIME}(n)$ NTM for $\mathrm{L}_{pad}$. Clearly, our NTM runs in non-deterministic time $cdt$, putting $\mathrm{L}$ inside $\mathrm{NTIME}(t)$.