Characterizing relations by forbidden induced subsets Working with relations in a purely set theoretic manner i.e. as just sets of ordered pairs, we see for any relation $R$ there exists unique inclusion minimal sets $A$ and $B$ such that $R\subseteq A\times B$ so that if we write $\text{dom}(R)=A$ and $\text{rng}(R)=B$ with $\text{fld}(R)=A\cup B$ then $\text{fld}(R)$ by definition is the unique inclusion minimal set $X$ such that $R\subseteq X\times X$. Now we say any relation $L$ is isomorphic to the relation $R$ and write $R\cong L$ if and only if there exists a bijection $f:\text{fld}(R)\to \text{fld}(L)$ such that $\forall x,y\in\text{fld}(R)\left(xRy\iff f(x)Lf(y)\right)$ further we say any relation $J$ is an induced subset of $R$ if and only if there exists a set $S$ such that we have $J=R\cap S^2$. With that said, note if we write:
$$R\text{ is reflexive}\iff \forall a\in\text{fld}(R)\left(aRa\right)$$
$$R\text{ is total}\iff \forall a,b\in\text{fld}(R)\left(aRb\lor bRa\right)$$
$$R\text{ is symmetric}\iff \forall a,b\in\text{fld}(R)\left(aRb\implies bRa\right)$$
Then if we let $\small A_{1}=\{(1,1),(1,2),(2,2)\}$, $\small A_{2}=\{(1,2),(2,1),(1,1)\}$, $\small A_{3}=\{(1,1),(2,1)\}$, as well as $\small A_{4}=\{(1,2),(2,1)\}$, $\small A_{5}=\{(1,1),(2,2)\}$, $\small A_{6}=\{(1,1),(1,2)\}$ and $\small A_{7}=\{(1,2)\}$ then we can prove:
$$\small R\text{ is reflexive}\iff \text{No induced subset of }R\text{ is isomorphic to a relation in }\{A_{2},A_{3},A_{4},A_{7},A_{6}\}$$
$$\small R\text{ is total}\iff \text{No induced subset of }R\text{ is isomorphic to a relation in }\{A_{2},A_{3},A_{4},A_{5},A_{6},A_{7}\}$$
$$\small R\text{ is symmetric}\iff \text{No induced subset of }R\text{ is isomorphic to a relation in }\{A_1,A_{3},A_{6},A_{7}\}$$
Further for many other relational properties such finite families of finite relations exist which act as "obstruction" sets, for example if we write $\small R\text{ is transitive}\iff \forall a,b,c\in\text{fld}(R)\left(aRb\land bRc\to aRc\right)$ then there is a finite family $\mathcal{F}$ of finite relations such that $R$ is transitive iff no induced subset of $R$ is isomorphic to a relation in $\mathcal{F}$ in fact the smallest such set $\mathcal{F}$ contains $34$ elements. Now it seems any relational property that can be characterized with a universal quantifier over its field of elements and any first order sentence identifying the elements among each other or as pairs in a relation, must always have some finite family of finite relations which act as an obstruction set as previously defined (and vice versa). Or more formally if we call any property $\mathcal{P}$ universal iff there exists $n\in\mathbb{N}$ such that there is a fixed formula $\phi_R(x_1,\ldots x_n)$ formed using only the conjugation, disjunction, or negation of those propositions of the form $x_i=x_j$ or $x_iRx_j$ for any relation $R$ and any integers $1\leq i,j\leq n$ such that for any relation $A$ we have: $\small (A\text{ has property }\mathcal{P})\iff \forall a_1,\ldots a_n\in\text{fld}(A)\left[\phi_A(a_1,\ldots a_n)\right]$ and if we also say any family of relations $\mathcal{F}$ obstructs $\mathcal{P}$ when any relation $R$ has property $\mathcal{P}$ iff no induced subset of $R$ is isomorphic to an element of $\mathcal{F}$, then I am claiming for any property $\mathcal{P}$ that:
$$\small\mathcal{P}\text{ is universal }\iff \text{There exists a finite family of finite relations which obstructs }\mathcal{P}$$
It seems like it should be relatively easy to prove and I have an argument in my mind, but I'm not able to write up anything that isn't horrible looking. I sense that I lack some pretty basic knowledge in logic/language theory which would allow me to write up a formal proof. What would one look like?
 A: I am going to rephrase the question in terms of first order relational structures. I believe the answer will be sufficiently close to what you are looking for.
Let $L$ be the first order language containing only a binary relation symbol $R$. Call a class $\mathcal{P}$ of $L$-structures universal if there is a collection $T$ of universal $L$-sentences such that $M\in\mathcal{P}$ if and only if $M\models T$. Call $\mathcal{P}$ strongly universal if, moreover, $T$ is finite--equivalently, there there is a single universal $L$-sentence $\phi$ such that $M\in\mathcal{P}$ if and only if $M\models \phi$. (So what I am calling strongly universal is the version of what you are calling universal.)

Theorem. Let $\mathcal{P}$ be a class of $L$-structures.
$(a)$ $\mathcal{P}$ is universal if and only if there is a class $\mathcal{F}$ of finite $L$-structures such that $M\in\mathcal{P}$ if and only if $M$ omits every element of $\mathcal{F}$ as a (induced) substructure. 
$(b)$ $\mathcal{P}$ is strongly universal if and only if there is a finite class $\mathcal{F}$ of finite $L$-structures such that $M\in\mathcal{P}$ if and only if $M$ omits every element of $\mathcal{F}$ as a (induced) substructure.
Proof. Suppose first that there is a class $\mathcal{F}$ of finite $L$-structures such that $M\in\mathcal{P}$ if and only if $M$ omits every element of $\mathcal{F}$. Given $A\in\mathcal{F}$, let $\psi_A$ be a universal $L$-sentence that holds in an $L$-structure if and only if it omits $A$ (this is possible since $A$ is finite). Let $T=\{\psi_A:A\in\mathcal{F}\}$. Then $M\in\mathcal{P}$ if and only if $M\models T$. So this proves the right-to-left direction of both statements. 
Now we prove the left-to-right direction of $(a)$. Suppose $\mathcal{P}$ is universal, witnessed by the $L$-theory $T$. Since every sentence in $T$ is universal, it follows that $\mathcal{P}$ is closed under substructures (i.e., if $M\in\mathcal{P}$ and $N$ is a substructure of $M$ then $N\in\mathcal{P}$). Now let $\mathcal{F}$ be the class of all finite $L$-structures not in $\mathcal{P}$. We claim that an $L$-structure $M$ is in $\mathcal{P}$ if and only if $M$ omits every element of $\mathcal{F}$. For one direction, if $M$ is in $\mathcal{P}$ then $M$ omits $\mathcal{F}$ by definition of $\mathcal{F}^*$ and the fact that $\mathcal{P}$ is closed under substructures. For the other direction, suppose $M$ is an $L$-structure not in $\mathcal{P}$. Then $M\models\neg\phi$ for some $\phi\in T$. Since $\phi$ is universal and $L$ is relational, there is some finite substructure $A$ of $M$ such that $A\models\neg\phi$. So $A\not\in\mathcal{P}$, which implies $A\in\mathcal{F}$. So $M$ does not omit $\mathcal{F}$. 
Finally, for the left-to-right direction of $(b)$, suppose $\mathcal{P}$ is strongly universal, witnessed by the $L$-sentence $\phi$. Let $\mathcal{F}$ be given as in part $(a)$. We claim that there is some finite $\mathcal{F}_0\subseteq\mathcal{F}$ such that an $L$-structure $M$ omits $\mathcal{F}$ if and only if it omits $\mathcal{F}_0$. For each $A\in\mathcal{F}$, let $\psi_A$ be the same $L$-sentence as before. Let $T=\{\psi_A:A\in\mathcal{F}\}$. Then $M\in\mathcal{P}$ if and only if $M\models T$. So $T\models \phi$. By the compactness theorem, there is some finite $T_0\subseteq T$ such that $T_0\models\phi$. So if $\mathcal{F}_0=\{A\in\mathcal{F}:\psi_A\in T_0\}$, then $\mathcal{F}_0$ is as desired. 

Remark 0. As Emil Je&rcaron;ábek points out in the comments, the use of compactness is overkill in $(b)$. One can instead just list out the structures witnessing a failure of $\phi$. 
Remark 1. In the setup of your question, it seems like you might have allowed the formula $\phi_R$ to depend on the concrete relation $R$ (depending on what you mean with the word "fixed"). But notice that in all of the examples (reflexive, symmetric, etc.) the formula is uniform in $R$. This is reflected in the notion of strongly universal by having the sentence $\phi$ depend only on the class $\mathcal{P}$. This is necessary. Otherwise you could form a class $\mathcal{P}$ such that $\phi_R$ is $\forall x(x=x)$ if $\text{fld}(R)$ is finite, and $\phi_R$ is $\forall x(x\neq x)$ if $\text{fld}(R)$ is infinite. So $\mathcal{P}$ consists just of the relations with finite universes, and thus cannot be characterized by omitting a class of finite structures (let alone a finite class). 
