Let $P=(p_{ij})$ and $Q=(q_{ij})$ be the $n\times n$ transition matrices for the two respective Markov chains, where $n\ge2$. Your conditions imply the following:
\begin{gather}p_{nn}=q_{nn},\\
p_{nj}<q_{nj}\text{ if }1\le j\le n-1.
\end{gather}
Hence, $1=\sum_{j=1}^n p_{nj}<\sum_{j=1}^n q_{nj}=1$, which is a contradiction (which makes any conclusion whatsoever true).
---
**Added:** In a comment, the OP suggested that the conjecture be modified by now assuming $n$ to be an absorbing state as well, so that the conditions become
\begin{gather}p_{11}=q_{11}=p_{nn}=q_{nn}=1,\\
p_{ii}=q_{ii}\text{ if }1\le i\le n, \\
p_{ij}q_{ij}>0\text{ if }1<i<n,\\ p_{ij}>q_{ij}\text{ if }1<i<j\le n,\\
p_{ij}<q_{ij}\text{ if }1\le j<i<n.
\end{gather}
The conjecture then becomes that
\begin{gather}f_{P;ij}>f_{Q;ij}\text{ if }1<i<j\le n,\\
f_{P;ij}<f_{Q;ij}\text{ if }1\le j<i<n,
\end{gather}
where $f_{P;ij}$ is the probability that the first chain ever reaches $j$ from $i$, and $f_{Q;ij}$ is defined similarly.
This conjecture, too, is false in general. E.g., suppose that $n=5$,
$$P=\frac1{20}
\left(
\begin{array}{ccccc}
20 & 0 & 0 & 0 & 0 \\
4 & 4 & 4 & 4 & 4 \\
2 & 2 & 2 & 12 & 2 \\
4 & 4 & 4 & 4 & 4 \\
0 & 0 & 0 & 0 & 20 \\
\end{array}
\right),\quad
Q=\frac1{20}\left(
\begin{array}{ccccc}
20 & 0 & 0 & 0 & 0 \\
7 & 4 & 3 & 3 & 3 \\
12 & 4 & 2 & 1 & 1 \\
5 & 5 & 5 & 4 & 1 \\
0 & 0 & 0 & 0 & 20 \\
\end{array}
\right).
$$
Then
$$f_{P;32}=\frac13\not<\frac{69}{283}=f_{Q;32}.$$
**Added more:** This is in response to a request by the OP to provide details. Fix any $j\in[n]:=\{1,\dots,n\}$. Let $g$ denote the $j$th column of $P$ and let $R$ denote the matrix obtained from matrix $P$ by replacing the $j$th column in $P$ by the zero column. Then for the column matrix $f:=(f_{P;ij}\colon i\in[n])^T$ we have
$$f=\sum_{n=0}^\infty (R^n g);
$$
cf. e.g. the 5th display on page 85 and formula (2.6.3') on page 90 in [Resnick][1]. If all entries of the column matrix $g$ are nonzero, then the norm $\|R\|_{\infty,\infty}$ of the matrix $R$ considered as a linear operator from $\ell_n^\infty$ to $\ell_n^\infty$ will be $<1$, whence we will have $f=(\sum_{n=0}^\infty R^n)g=(I-R)^{-1}g$. However, in our case, because we have two absorbing states, every column of $P$ must have a zero entry. This difficulty is easy to circumvent, as follows. By the monotone convergence theorem, for $t\in(0,1)$
$$(I-tR)^{-1}g=\sum_{n=0}^\infty (tR)^n g\;\Big\uparrow\;\sum_{n=0}^\infty (R^n g)=f
$$
as $t\uparrow1$. So,
$$f=\lim_{t\uparrow1}(I-tR)^{-1}g.
$$
Since $(I-tR)^{-1}g$ is rational in $t$, the latter limit is easy to compute.
In particular, for $n=5$ we get
$$f_{P;32}=\frac {p_{34}p_{42}+p_{32}(1-p_{44})}
{(1-p_{33}) (1-p_{44})-p_{34} p_{43}},
$$
with the corresponding expression for $f_{Q;32}$. We see that $f_{P;32}$ is increasing in $p_{34}$. Also, one of the OP's conditions is $p_{34}>q_{34}$. So, to get $f_{P;32}>f_{Q;32}$, we should try to make $p_{34}$ large and $q_{34}$ small. At the same time, we may choose $p_{42}$, $p_{32}$, and $p_{43}$ close enough to $q_{42}$, $q_{32}$, and $q_{43}$ (respectively), also keeping in mind the conditions $p_{33}=q_{33}$ and $p_{44}=q_{44}$. This should (and does) result in $f_{P;32}>f_{Q;32}$, which disproves the conjecture.
**Added yet more:** The matter becomes much more transparent if we ignore, at least for a moment, the strictness of the inequality restrictions on the $p_{ij}$'s and $q_{ij}$'s. Then we may assume that $p_{34}=q_{31}=p_{42}=q_{42}=1$. Then clearly $f_{P;32}=1\not\le0=f_{Q;32}$. If you still insist on the strictness of the inequality restrictions, then you can have it by the continuity of $f_{P;32}$ on the appropriate domain.
[1]: https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=4&cad=rja&uact=8&ved=2ahUKEwjezNzWsdnlAhVERKwKHeK7CgsQFjADegQIBBAC&url=http%3A%2F%2Fwww.fcfm.buap.mx%2Fjzacarias%2Fcursos%2Fprocesos%2Flibros%2Fbook3pe.pdf&usg=AOvVaw2S9tF8Bs3QpzA0FLECZZkJ