The answer is yes.
Indeed, by rescaling, without loss of generality (wlog) $c=1$. To simplify the notations, let $f:=F$, $a:=a_n$, $b:=b_n$, $c:=c_n$, $t:=\lambda_n$, $D:=D_n$. Passing to a subsequence, wlog $a\to a_*\in[0,1]$ and $t\to t_*\in(0,1]$. Also, wlog $a+2\le c$, since $a\le1$ and $c\to\infty$. Also, wlog $b>c$, since wlog $t>0$ and $c>a$.
By the convexity of $f$ and inequalities $a+1\le a+2\le c$, \begin{equation*} f(a+1)\ge f(c)+\frac{a+1-c}{b-c}\,(f(b)-f(c)).\tag{1} \end{equation*} Using now the convexity of $f$ again together with the inequality $a+2\le c$ and (1), we have \begin{align*} 0\le d&:=\frac{f(a)+f(a+2)}2-f(a+1) \\ &\le \frac{f(a)}2+\frac12\,\frac{(c-a-2)f(a)+2f(c)}{c-a}-f(a+1) \\ &=\frac{(c-a-1)f(a)+f(c)}{c-a}-f(a+1) \\ &\le\frac{(c-a-1)f(a)+f(c)+D}{c-a} \\ &\ \ -\Big(f(c)+\frac{a+1-c}{b-c}\,(f(b)-f(c))\Big) \\ &=\frac{b-a-1}{b-a}\frac Dt\sim\frac D{t_*}\to0, \end{align*} so that \begin{equation*} d\to0. \tag{2} \end{equation*} On the other hand, \begin{equation*} d\to\frac{f(a_*)+f(a_*+2)}2-f(a_*+1)>0 \end{equation*} by the strict convexity of $f$. This contradicts (2). $\Box$
Remark: As seen from the above, condition $c\to\infty$ can be relaxed to $\liminf(c-a)>0$.