In general a $T^{eq}$ construction in continuous logic, as it's typically defined, can only possibly add hyperimaginaries corresponding to equivalence relations defined by countable partial types. If you consider the uncountable discrete theory $T$ with unary predicates $P_i$ for each $i<\omega_1$ where for each pair of disjoint finite sets $X,Y\subset \omega_1$ $T$ contains an axiom $\exists x\bigwedge_{i\in X}P_i(x)\wedge \bigwedge_{i\in Y} \neg P_i(x)$. Then if you look at the type definable equivalence relation $E(x,y)=\{P_i(x) \leftrightarrow P_i(y) : i<\omega_1\}$ the hyperimaginary where you quotient out by $E$ is not equivalent to any continuous imaginary of $T$, because any continuous formula is definable over some countable set of $P_i$'s.


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EDIT: It's actually not even true for countable theories. Here is a counterexample.

Let $\mathcal{L}=\{P_i\}_{i<\omega}$ be a countable sequence of binary predicates and let $\Sigma$ be the $\mathcal{L}$-theory consisting of the following for each $i<\omega$:

 - $\forall x P_i(x,x)$
 - $\forall xy(P_i(x,y)\leftrightarrow P_i(y,x))$
 - $\forall xyz(P_{i+1}(x,y)\wedge P_{i+1}(y,z) \rightarrow P_i(x,z))$

Note that these imply $\forall xy(P_{i+1}(x,y) \rightarrow P_{i}(x,y))$. 

Let $E(x,y)$ be the partial type given by the formulas $\{P_i(x,y)\}_{i<\omega}$. It's clear that in any model of $\Sigma$, $E$ gives a type-definable equivalence relation.

Let $\mathcal{K}$ be the class of finite models of $\Sigma$. First of all I claim that this class contains only countably many isomorphism types. To see this note that for any $\mathfrak{A} \in \mathcal{K}$ and any $a,b\in \mathfrak{A}$, the quantifier free type of $ab$ is uniquely determined by the quantity $Q(x,y)=2^{-i}$ where $\mathfrak{A} \models P_i(x,y)\wedge \neg P_{i+1}(x,y)$ or $Q(x,y)=0$ if $\mathfrak{A} \models E(x,y)$. This has countably many values and the isomorphism type of an model of $\mathcal{K}$ is uniquely determined by the values of $Q$ on its pairs of elements.

Also since $\Sigma$ is a universal theory $\mathcal{K}$ is closed under substructures. 

I claim that $\mathcal{K}$ has the amalgamation property. To see this let $\mathfrak{A}\subset \mathfrak{B}$ and $\mathfrak{A} \subset \mathfrak{C}$ with $\mathfrak{A}$, $\mathfrak{B}$, and $\mathfrak{C}$ all models of $\Sigma$. We can give an amalgamation $\mathfrak{D}$ whose universe is $B\cup C$ and where for any $b \in B \setminus A$ and $c \in C \setminus A$ we set $Q(b,c) = \min(1,2 \min_{a \in A}\max(Q(b,a),Q(a,c)))$. In other words we introduce only the $P_i$ relationships that are absolutely necessary according to $\Sigma$. 

So we have that $\mathcal{K}$ is a Fraïssé class and we can find a Fraïssé limit $\mathfrak{F}$. Let $T=\mathrm{Th}(\mathfrak{F})$. Since the automorphism type of every finite tuple in $\mathfrak{F}$ is entirely determined by its quantifier free type and since every quantifier free type consistent with $T$ occurs in $\mathfrak{F}$, $T$ admits quantifier elimination. 

Let $\mathcal{L}^\prime$ be a continuous language with a single binary $[0,1]$-valued predicate symbol $Q$. Interpret $\mathfrak{F}$ as an $\mathcal{L}^\prime$-structure $\mathfrak{F}^\prime$ using the definition of $Q$ given above. It's clear that $\mathfrak{F}$ and $\mathfrak{F}^\prime$ are interdefinable and more than that interdefinable in a quantifier free way. In particular I claim that $T^\prime = \mathrm{Th}(\mathfrak{F}^\prime)$ eliminates quantifiers in the sense of continuous logic. In particular this implies that the $2$-type of any pair $ab$ is entirely determined by the value of $Q(a,b)$. Also note that by construction we have that $\mathfrak{F} \models E(a,b)$ if and only if $\mathfrak{F}^\prime \models Q(a,b)$ (i.e. that it is $0$).

Let $\rho(x,y)$ be a $\varnothing$-definable $[0,1]$-valued pseudo-metric such that $\mathfrak{F} \models E(a,b)$ implies $\mathfrak{F}^\prime  \models \rho(a,b)$. By quantifier elimination there must be some continuous function $\alpha:[0,1]\rightarrow[0,1]$ such that $\rho(x,y)=\alpha(Q(x,y))$. Now for each $i<\omega$ consider the finite $\mathcal{L}$-structure $\mathfrak{A}$ with $A=\{a,b,c\}$ such that $Q(a,b)=0$, $Q(a,c)=2^{-i}$, and $Q(b,c)=2^{-i-1}$. Note that the third axiom schema for $\Sigma$ translates to $Q(x,z)\leq 2 \max(Q(x,y),Q(y,z))$, so in particular $\mathfrak{A}$ is a model of $\Sigma$. Therefore it embeds into $\mathfrak{F}$ with some map $f:\mathfrak{A}\rightarrow \mathfrak{F}$. Since $Q(f(a),f(b))=0$ we must have that $\rho(f(a),f(b))=0$. Since $\rho$ is a pseudo-metric this implies that $\rho(f(a),f(c))=\rho(f(b),f(c))$. Therefore we have that $\alpha(2^{-i})=\alpha(2^{-i-1})$. Since this is true for any $i<\omega$, we have that $\alpha$ is constant on the set $\{2^{-i}\}_{i<\omega}$. It must be the case that $\alpha(0)=0$ since $\mathfrak{F} \models E(a,b)$ implies $\mathfrak{F}^\prime  \models \rho(a,b)$, therefore by continuity $\alpha(r)=0$ for all $r\in[0,1]$ and $\rho$ is the trivial pseudo-metric.

So we have a theory with a countably type-definable equivalence relation (definable over $\varnothing$) in which the only $\varnothing$-definable pseudo-metric that is at least as coarse as it is the trivial pseudo-metric.

I'm not sure if allowing more parameters for the pseudo-metric can save it but I'm not very hopeful.