I am assuming $t$ is real. Since
$$ \left|\frac{(-1)^{n-1}}{n^{1/2+it}}+\frac{(-1)^{n+1-1}}{(n+1)^{1/2+it}} \right|
\le|1/2+it|\max_{0\le s\le1}\left|\frac1{(n+s)^{3/2+it}}\right|
=\sqrt{t^2+1/4}\,\frac1{n^{3/2}},
$$
the series $\sum_{1}^\infty\frac{(-1)^{n-1}}{n^{1/2+it}}$ converges, whereas the harmonic series diverges. So indeed, for each real $t$ there is some natural $N(t)$ such that the inequality in question holds for all natural $N\ge N(t)$.

**Addition in response to the modification of the question:** More specifically,
$$\left|\sum_{1}^N\frac{(-1)^{n-1}}{n^{1/2+it}}\right|
\le1+\sqrt{t^2+1/4}\,\sum_{j=1}^\infty\frac1{(2j-1)^{3/2}}
=1+c\sqrt{t^2+1/4},$$
where $c:=(1-\frac1{2\sqrt2})\zeta(3/2)<1.7$, whereas $H_N>\gamma+\ln N$. So, the inequality in question holds for all natural $N\ge N(t)$, where
$N(t):=\exp\{1-\gamma+c\sqrt{t^2+1/4}\}$ and $\gamma$ is the Euler constant.

**Another addition: Better bounds** Luca asked if the above bound on $N(t)$ can be improved, to get something closer to $\lceil t\rceil^2 $.

**1.** One such improvement is an easy modification of the above reasoning. Indeed, let $t\to\infty$ and let $N>A:=\lceil t\rceil$. Then
\begin{equation*}
\left|\sum_{1}^N\frac{(-1)^{n-1}}{n^{1/2+it}}\right|
\le\left|\sum_{1}^A\frac{1}{n^{1/2}}\right|
+\left|\sum_{A+1}^N\frac{(-1)^{n-1}}{n^{1/2+it}}\right|
\ll A^{1/2}+\frac{\sqrt{t^2+1/4}}{A^{1/2}}\sim 2t^{1/2}.
\end{equation*}
So, the inequality in question holds for all natural $N\ge N(t)$, where
$N(t):=\exp\{c_1+c_2 t^{1/2}\}$, where $c_1$ and $c_2$ are some positive real constants.

Further improvements seem very difficult:

**2.** Again, let $t\to+\infty$. The Lindelöf hypothesis -- which is implied by the Riemann hypothesis (\url{https://en.wikipedia.org/wiki/Lindel%C3%B6f_hypothesis},
\url{https://terrytao.wordpress.com/tag/riemann-zeta-function/}) -- states that
$|\zeta(1/2+it)|\ll t^\epsilon$ for any $\epsilon>0$.
Since $\eta(z):=\sum_{1}^\infty\frac{(-1)^{n-1}}{n^z}=\zeta(z)(1-2^{1-z})$, the Lindelöf hypothesis implies $|\eta(1/2+it)|\ll t^\epsilon$ for any $\epsilon>0$. So, assuming $N>t^{2-2\epsilon}$, we have
\begin{equation*}
\left|\sum_{1}^N\frac{(-1)^{n-1}}{n^{1/2+it}}\right|
\le\left|\sum_{1}^\infty\frac{(-1)^{n-1}}{n^{1/2+it}}\right|
+\left|\sum_{N+1}^\infty\frac{(-1)^{n-1}}{n^{1/2+it}}\right|
\ll t^\epsilon+\frac{\sqrt{t^2+1/4}}{N^{1/2}}\ll t^\epsilon.
\end{equation*}
So, modulo the Lindelöf hypothesis, the inequality in question holds for all natural $N\ge N(t)$, where
$N(t):=\exp\{c_1+c_2 t^\epsilon\}$, where $c_1$ and $c_2$ are some positive real constants.

**3.** By plotting with Mathematica, I have not been able to find a counterexample
to the inequality
\begin{equation*}
|\eta(1/2+it)|\le2\ln t \tag{*}
\end{equation*}
for any $t\in[2,10^7]$; the closest to $2$ value of $|\eta(1/2+it)|/\ln t$ for such $t$ seems to occur near $t=6.5\times10^5$. \
One may conjecture that $(*)$ holds for all real $t\ge2$.

Again, let $t\to+\infty$. Let $c_*=0.86568\dots$ be the positive root of the equation $c_*^2 e^{c_*-\gamma}=1$, and let $N(t):=bt^2$, where $b$ is any fixed real number greater than $e^{c_*-\gamma}=1/c_*^2=1.3343\dots$, so that $b^{-1/2}<c_*$.
Assume that $(*)$ holds for all large enough real $t$. Then for all large enough $t>0$ and all $N\ge N(t)$ we have
\begin{equation*}
\left|\sum_{1}^N\frac{(-1)^{n-1}}{n^{1/2+it}}\right|
\le\left|\sum_{1}^\infty\frac{(-1)^{n-1}}{n^{1/2+it}}\right|
+\left|\sum_{N+1}^\infty\frac{(-1)^{n-1}}{n^{1/2+it}}\right|
\le2\ln t+\frac{\sqrt{t^2+1/4}}{N^{1/2}}\,(1+o(1))
\end{equation*}
\begin{equation*}
\le2\ln t+\frac{1+o(1)}{b^{1/2}}
\le2\ln t+c_*
<
\ln N+\gamma<H_N.
\end{equation*}
So, modulo conjecture $(*)$, the inequality in question holds for all natural $N\ge N(t):=1.3344t^2$.

This is still a bit short of $N(t)=\lceil t\rceil^2$. However, this may to an extent explain that "empirical" bound, $N(t)=\lceil t\rceil^2$.

However, conjecture $(*)$ is not true: according to \url{https://arxiv.org/abs/1704.06158v2}, $\limsup_{t\to\infty}|\eta(1/2+it)|/e^{c\ell(t)}>1$ for any $c\in(0,1)$, where $\ell(t):=\sqrt{\ln t\;\ln\ln\ln t\,/\,\ln\ln t}$.

It follows that necessarily $N(t)>\exp\{e^{(1-o(1))\ell(t)}\}$ for some however large $t$, and of course $\exp\{e^{(1-o(1))\ell(t)}\}$ is much greater than $t^2$ for large $t>0$. Indeed, take any $c\in(0,1)$. Then, for some however large $t$ and any natural $N\in(t^2,\exp\{e^{c\ell(t)}-3\})$,

\begin{equation*}
H_N\le1+\ln N<e^{c\ell(t)}-2<|\eta(1/2+it)|-2<|\eta(1/2+it)|-2\frac t{N^{1/2}}
\end{equation*}
\begin{equation*}
\le\left|\sum_{1}^\infty\frac{(-1)^{n-1}}{n^{1/2+it}}\right|
-\left|\sum_{N+1}^\infty\frac{(-1)^{n-1}}{n^{1/2+it}}\right|
\le \left|\sum_{1}^N\frac{(-1)^{n-1}}{n^{1/2+it}}\right|,
\end{equation*}
so that the inequality in question fails to hold.