$ \newcommand{\Hom}{\mathrm{Hom}} \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}$I have a construction that works in some cases. It is correct as long as all categories involved have filtered colimits and the right adjoints commute with them. If this answer is hard to understand, it might be helpful to do what I did when thinking of it and view the span in the category of adjunctions as consisting of the free/forgetful adjunctions $\mathbf{Set} \to \mathbf{Ring}$ and $\mathbf{Set} \to \mathbf{Group}$. Then the final category we are constructing is just sets with unrelated group and ring structures, and the main difficulty is categorically defining the free adjoint to the forgetful functor from ring-groups to rings or groups.
Let $F_1: \mathcal C \to \mathcal D_1$ and $G_1: \mathcal D_1 \to \mathcal C$ be one adjunction, and $F_2: \mathcal C \to \mathcal D_2$ and $G_2: \mathcal D_2 \to \mathcal C$ be the other.
Let $\mathcal B$ be the category of pairs of objects $X \in D_1, Y \in D_2$ with an isomorphism between $G_1 X$ and $G_2 Y$. There are natural projection functors $P_1: \mathcal B \to \mathcal D_1$ and $P_2: \mathcal B \to \mathcal D_2$. We will construct adjoints $Q_1: \mathcal D_1 \to \mathcal B$ and $Q_2 : \mathcal D_2 \to \mathcal B$.
We need an adjunction of the form $\Hom_{\mathcal B} (Q_1 X_0, (X,Y) ) = \Hom_{\mathcal D_1} (X_0,X)$. We have a natural map:
$\Hom_{\mathcal D_1}(X_0,X) \to \Hom_{\mathcal C} ( G_1X_0,G_1X) = \Hom_{\mathcal C}(G_1X_0,G_2 Y) \to \Hom_{\mathcal D_2} (F_2G_1 X_0, F_2G_2 Y) \to \Hom_{\mathcal D_2} (F_2 G_1 X_0, Y)$
Let $Y_1 = F_2G_1 X_0$. Then by the same logic we have a natural map $\Hom_{\mathcal D_1}(X_0,X) \to \Hom_{\mathcal D_1} ( F_1 G_2 Y_1, X)$. But we have something slightly more refined. $F_1 G_2 Y_1 = F_ 1 G_2 F_2 G_1 X_0$. Every element of $\Hom_{\mathcal D_1}(X_0,X)$ gives a commutative diagram:
$$ \begin{array}{c} F_1G_1X_0 & \ra{} & X_0 \\ \da{} & & \da{} \\ F_1G_2F_2 G_1 X_0 & \ra{} & X \end{array} $$
Call $X_2$ the limit of the top-left three-quarters of this diagram. Every map $X_0 \to X$ gives a natural map $X_2 \to X$. Then let $Y_3$ be the limit of the corresponding diagram starting at $Y_1$. Every map $X_0 \to X$ gives a corresponding map $Y_3 \to Y$.
Define $Y_4$, $X_5$, etc., similarly. Let $X_{\infty} = \lim_{n\to\infty} X_n$ and let $Y_\infty = \lim_{n\to\infty} Y_n$. Then $G_1 X_\infty = G_2 Y_\infty$, because they are both equal to the limit of the sequence $G_1 X_0 \to G_2 Y_1 \to G_1 X_2 \to G_2 Y_3 \to \dots$, because right adjoints preserve limits and the limit of a subsequence is the limit of the sequence. So $(X_{\infty}, Y_{\infty})$ is an element of $\mathcal B$, and we have a natural map $\Hom _{\mathcal D_1}(X_0,X)= \Hom_ {\mathcal B} ( ( X_\infty, Y_\infty), (X,Y))$. To see this is a bijection note that an element on the right side is just a sequence of maps $X_n \to X$, and $Y_m \to Y$. We need to show that all these maps are determined by the map $X_0 \to X$. They satisfy the consistency condition that the families of map $G_1 X_n \to G_1 X$, $G_2 Y_2 \to G_2 Y$ agree.
So for instance we have the unit map $G_1 X_0 \to G_2 F_2 G_1 X_0 = G_2 Y_1$. The map $Y_1 \to Y$ is the unique map such that the functor $F_2$ applied with it, composed with this unit, is the map $G_1 X_0 \to G_1 X=G_2 Y$. In other words it is exactly the map $F_2 G_1 X_0 \to Y$ one obtains by the adjunction. Iterating the process, we see that the maps from $X_2$, $Y_3$, etc. are also so determined.
So define $Q_1 X_1 = (X_{\infty}, Y_{\infty})$. Define $Q_2$ similarly. This completes our span to a commutative square. Is it universal? It clearly is one way. If $\mathcal B'$ is any category with functors $P_1': \mathcal B' \to \mathcal D_1$ and $P_2': \mathcal B' \to \mathcal D_2$, we obtain a unique map $\mathcal B' \to \mathcal B$ by definition. So we only need to construct an adjoint of that map. We do this as follows. Given $X,Y$, and an isomorphism $G_1X = G_2Y$, we form in $\mathcal B'$ the span
$$ \begin{array}{c} Q_1' F_1G_1 X & \ra{} & Q_1' X \\ \da{} & \\ Q2' Y & \end{array} $$
with the bottom arrow coming from the identification $Q_1'F_1G_1 X = G_2' F_2 G_1 X = Q_2' F_2 G_2 Y$.
If the pushout of this span in $\mathcal B'$ exists, we can use it to construct an adjoint functor. A $\Hom$ from this to an element $Z \in B'$ consits of a $\Hom(Q_1' X, Z)$ plus a $\Hom(Q_2'Y, Z)$ that agree on $\Hom(Q_1'F_1 G_1 X, Z)$. By adjunction, this is the same as a $\Hom(X, P_1'Z)$ plus a $\Hom(Y, P_1'Z)$ that agree on $\Hom(G_1X, G_1 P_1'Z)$. This is exactly a an element of $\Hom_B((X,Y), (P_1'Z,P_2'Z))$.
So assuming all categories have the appropriate limits we are using and they are preserved in the right way this should work.
Edit: If the relevant categories do not have filtered colimits, this is false. Take $\mathcal C$ to be finite sets, $\mathcal D_1$ to be finite sets with an action of the finite group $G_1$, $\mathcal D_2$ to be finite sets with an action of the finite group $G_2$. The adjunctions are the forgetful/free adjunction, which exists because the group is finite. The pushout should equal finite sets with an action of the free product $G_1 * G_2$, but it cannot because the free functor takes you to infinite sets. If $H$ is any finite quotient of $G_1 * G_2$, then $H$-sets forms a commutative diagram in the category of adjunctions. If there were a universal adjunction then for $X \in \mathcal D_1$, $Q_1 P_1X$ would be at least as large as the free objects in any of these categories. But as $H$ grows, the free objects grow arbitrarily large, so this is impossible.
